Topic 10 Sequences and Series

10.1 Sequences

A sequence $\left\{a_n\right\}$ is a list of numbers written in a definite order. It can be considered as a function whose domain are (positive) integers.

The limit $\underset{n\to \infty }{lim}{a}_n$ of a sequence $\left\{a_n\right\}$ is defined similarly to the limit $\underset{n\to \infty }{lim}f\left(x\right)$ of the function $f$. Indeed, we have the following theorem.

Theorem 10.1 If $\underset{x\to \infty }{lim}f\left(x\right)=L$ and $f\left(n\right)=a_n$, then $\underset{n\to \infty }{lim}{a}_n=L$.

If the limit $\underset{n\to \infty }{lim}{a}_n$ exists, we say the sequence $\left\{a_n\right\}$ is convergent. Otherwise, we say the sequence $\left\{a_n\right\}$ is divergent.

Convergence sequences have the same rules and properties of limits as functions. For example, we may apply L’Hospital’s rule to $\frac{0}{0}$ or $\frac{\infty }{\infty }$ type sequences.

When determining the convergence of a sequence or finding the limit, we may not know the explicit expression of the sequence the following theorem may be useful.

Theorem 10.2 (Monotone Convergence Theorem) Let $\left\{a_n\right\}$ be a bounded sequence, that is $|a_n|<M$ for all $n$. If there exists an integer $N$ such that either $a_n\le a_{n+}$ for all $n>N$ or $a_n\ge a_{n+}$ for all $n>N$, that is the sequence is either increasing or decreasing (a.k.a. monotonic), then the sequence is convergent.

Exercise 10.1 List the first five terms of the sequence $$a_n=\frac{2n}{n^2+1}$$

Solution.

The first five terms are $$1,\frac{4}{5},\frac{3}{5},\frac{8}{17},\frac{5}{13}$$

Exercise 10.2 List the first five terms of the sequence $$a_1=2,a_2=1,a_{n+1}=a_n-a_{n-1}$$

Solution.

The first two terms have been given. Using the recurrence formula, we find $$a_3=-1,a_4=-2,a_5=-1.$$

Exercise 10.3 Find a formula for the general term $a_n$ of the following sequence, assuming that the pattern of the first few terms continues. $$\{\frac{1}{2},-\frac{4}{3},\frac{9}{4},-\frac{16}{5},\frac{25}{6},\dots \}$$

Solution.

The sequence can be explicitly expressed as $${\left\{\frac{(-1{)}^{n+1}{n}^2}{n+1}\right\}}_{n=1}^{\infty }$$

Exercise 10.4 Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_n=\frac{3+5n^2}{n+n^2}$$

Solution.

For such a rational expression, to find the limit, we divide the highest degree monomial from both the numerator and the denominator and apply the quotient (or product) rule of limits. $$\underset{n\to \infty }{lim}\frac{3+5n^2}{n+n^2}=\underset{n\to \infty }{lim}\frac{\frac{3}{n^2}+5}{\frac{1}{n}+1}=\frac{\underset{n\to \infty }{lim}\left(\frac{3}{n^2}\right)+5}{\underset{n\to \infty }{lim}\left(\frac{1}{n}\right)+1}=5.$$ The sequence converges to $5$.

Exercise 10.5 Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_n=\frac{3^{n+2}}{5^n}$$

Solution.

We rewrite the expression in the form $a\cdot r^n$ form and use the fact that $\underset{n\to \infty }{lim}{r}^n=0$ if $|r|<1$ to determine the convergence.

$$\underset{n\to \infty }{lim}\frac{3^{n+2}}{5^n}=9\underset{n\to \infty }{lim}{\left(\frac{3}{5}\right)}^n=9\cdot 0=0.$$ The sequence converges to $0$.

Exercise 10.6 Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_n=e^{2n/(n+2)}$$

Solution.

Let $f\left(x\right)=e^{2x/(x+2)}$. Then $f$ is continuous whenever $x>0$. We may use the theorem that $$\underset{n\to \infty }{lim}{e}^{2n/(n+2)}=\underset{x\to \infty }{lim}{e}^{2x/(x+2)}$$ to determine the convergence.

Use the fact that $y=e^x$ is continuous and $\underset{x\to \infty }{lim}\left(2x/\right(x+2\left)\right)=2$, we get $$\underset{x\to \infty }{lim}{e}^{2x/(x+2)}=e^{\underset{x\to \infty }{lim}\left(2x/\right(x+2\left)\right)}=e^2.$$ The sequence converges to $e^2$.

Exercise 10.7 Determine whether the sequence converges or diverges. If it converges, find the limit. $$\left\{\frac{(2n-1)!}{(2n+1)!}\right\}$$

Solution.

Recall that $$n!=n\cdot (n-1)\cdot (n-2)\cdot \cdots \cdot 2\cdot 1.$$

Then $$\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)\cdot (2n-2)\cdot \cdots \cdot 2\cdot 1}{(2n+1)\cdot 2n\cdot (2n-1)\cdot (2n-2)\cdot \cdots \cdot 2\cdot 1}=\frac{1}{(2n+1)\cdot 2n}$$ Therefore, $$\underset{n\to \infty }{lim}\left\{\frac{(2n-1)!}{(2n+1)!}\right\}=\underset{x\to \infty }{lim}\frac{1}{(2n+1)\cdot 2n}=0.$$

The sequence converges to $0$.

Exercise 10.8 Determine whether the sequence converges or diverges. If it converges, find the limit. $$\left\{n^2{e}^{-n}\right\}$$

Solution.

Let $f\left(x\right)=x^2{e}^{-x}$. Then $f$ is continuous and $$\begin{array}{cc}\underset{x\to \infty }{lim}{n}^2{e}^{-n}=& \underset{x\to \infty }{lim}{x}^2{e}^{-x}\\ =& \underset{x\to \infty }{lim}\frac{x^2}{e^x}\\ =& \underset{x\to \infty }{lim}\frac{2x}{e^x}\text{L'Hospital's Rule Applied}\\ =& \underset{x\to \infty }{lim}\frac{2}{e^x}\text{L'Hospital's Rule Applied}\\ =& 0.\end{array}$$ The sequence converges to $0$.

Exercise 10.9 Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_n=n\sin \left(1/n\right)$$

Solution.

Let $f\left(x\right)=x\sin \left(1/x\right)$. Then $f$ is continuous whenever $x>0$. Therefore, $$\begin{array}{cc}\underset{x\to \infty }{lim}n\sin \left(1/n\right)=& \underset{x\to \infty }{lim}x\sin \left(1/x\right)\\ =& \underset{x\to \infty }{lim}\frac{\sin \left(1/x\right)}{1/x}\\ =& \underset{t\to 0^{+}}{lim}\frac{\sin t}{t}t=1/x\\ =& \underset{t\to 0^{+}}{lim}\frac{\cos t}{1}\text{L'Hospital's Rule Applied}\\ =& 1.\end{array}$$ The sequence converges to $1$.

Exercise 10.10 Determine whether the sequence converges or diverges. If it converges, find the limit. $a_1=3$ and $a_{n+1}=\frac{a_n}{2}+1$ for all $n\ge 2$

Solution.

By the recurrence formula, we get $$a_1=3,a_2=\frac{5}{2},a_3=\frac{9}{4},a_4=\frac{17}{8}.$$

From those terms, we may claim that the sequence is decreasing and $a_n\ge 2$.

We prove the claim that $a_n\ge 2$ by induction. The monotonicity follows from this claim.

Clearly, $a_1\ge 2$. Suppose that the $a_n>2$. Then $$a_{n+1}=\frac{a_n}{2}+1\ge 1+1=2.$$

By the Monotone Convergence Theorem, $\underset{n\to \infty }{lim}{a}_n=L$ is a finite number.

Taking limits of both side of the equality $a_{n+1}=\frac{a_n}{2}+1$, we get $L=\frac{L}{2}+1$. Solve the equation for $L$, we get $$\underset{n\to \infty }{lim}{a}_n=L=2.$$

Remark: Mathematical induction is often used in dealing with recursive sequences.
A mathematical induction starts with checking the truth of a claim in the base case $k=n_0$, followed by a inductive step. In the inductive step, we assume the claim is true for $k=n$ (or $k\le n$), and then prove that the claim is true for $k=n+1$.

Exercise 10.11 Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_1=1a_{n+1}=3-\frac{1}{a_n}$$

Solution.

By the recurrence formula, we get $$a_1=1,a_2=2,a_3=3-\frac{1}{2}=\frac{5}{2},a_4=3-\frac{2}{5}=\frac{13}{5}.$$

From those terms, we may claim that the sequence is increasing and $1\le a_n\le 3$.

To prove that the sequence is increasing, we need to show that $$a_{n+1}-a_n=3-\frac{1}{a_n}-a_n=\frac{-a_n^2+3a_n-1}{a_n}>0.$$ If we can prove that $a_n\le \frac{3+\sqrt{5}}{2}$, then $a_{n+1}-a_n>0$. Because $-a_n^2+3a_n-1>0$ when $1\le a_n\le \frac{3+\sqrt{5}}{2}$.

This can be done by induction.

Clearly, $1\le a_1\le \frac{3+\sqrt{5}}{2}$ Suppose that the $1\le a_n\le \frac{3+\sqrt{5}}{2}$. Then $$1<a_{n+1}=3-\frac{1}{a_n}\le 3-\frac{2}{3+\sqrt{5}}=\frac{3+\sqrt{5}}{2}.$$

By the Monotone Convergence Theorem, $\underset{n\to \infty }{lim}{a}_n=L$ is a finite number.

Taking limits of both side of the equality $a_{n+1}=3-\frac{1}{a_n}$, we get $L=3-\frac{1}{L}$. Solve the equation, we get $$\underset{n\to \infty }{lim}{a}_n=L=\frac{3+\sqrt{5}}{2}.$$

10.2 Series

A series $\sum _{n=1}^{\infty }{a}_n$ is the (formal) sum of all terms of a sequence ${\left\{a_n\right\}}_{n=1}^{\infty }$. The sum $$s_n=a_1+a_2+a_3+\cdots +a_n=\sum _{i=1}^n{a}_i$$ is called a partial sum of the series.

A series $\sum _{n=1}^{\infty }{a}_n$ is convergent (divergent) if the sequence of partial sums $\left\{s_n\right\}$ is convergent (respectively, divergent).

Theorem 10.3 If a series $\sum _{n=1}^{\infty }{a}_n$ is convergent then $\underset{n\to \infty }{lim}{a}_n=0$. Conversely, if $\underset{n\to \infty }{lim}{a}_n\ne 0$, then $\sum _{n=1}^{\infty }{a}_n$ is divergent.

For convergence series, their linear combinations are also convergent.

Theorem 10.4 If $\sum _{n=1}^{\infty }{a}_n$ and $\sum _{n=1}^{\infty }{b}_n$ are convergence sequences, then $$\sum _{n=1}^{\infty }(c\cdot a_n+d\cdot b_n)=c\sum _{n=1}^{\infty }{a}_n+d\sum _{n=1}^{\infty }{b}_n,$$ where $c$ and $d$ are real numbers.

Using the formula $$1+r+r^2+r^3+\cdots +r^{n-1}=\frac{1-r^n}{1-r}$$ we know that the geometric series $$\sum _{n=0}^{\infty }a\cdot r^n=\frac{a}{1-r}\text{if}|r|<1.$$ Otherwise, it diverges.

Remark. When finding the sum of the geometric series, one should rewrite the series is in the following form $$a(1+r+r^2+r^3+\cdots )$$ before applying the above formula.

Exercise 10.12 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$4+3+\frac{9}{4}+\frac{27}{16}+\cdots$$

Solution.

The series can be expressed as $$4\cdot (1+\frac{3}{4}+{\left(\frac{3}{4}\right)}^2+{\left(\frac{3}{4}\right)}^3+\cdots )$$ Which is a scalar multiple of the geometric series with $r=\frac{3}{4}<1$.

Therefore, the series converges and $$4+3+\frac{9}{4}+\frac{27}{16}+\cdots =\frac{4}{1-\frac{3}{4}}=16.$$

Exercise 10.13 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum _{n=1}^{\infty }\frac{(-3{)}^{n-1}}{4^n}$$

Solution.

The series can be expressed as $$\frac{1}{3}\cdot (1+(-\frac{3}{4})+{(-\frac{3}{4})}^2+{(-\frac{3}{4})}^3+\cdots ).$$

Therefore, the series converges and $$\sum _{n=1}^{\infty }\frac{(-3{)}^{n-1}}{4^n}=\frac{\frac{1}{4}}{1-(-\frac{3}{4})}=\frac{1}{7}.$$

Exercise 10.14 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\frac{1}{15}+\cdots$$

Solution.

The series can be expressed as $$\frac{1}{4}\cdot (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots ).$$ which is a multiple of the harmonic series.

Because the harmonic series diverse, the series in this question diverges.

Exercise 10.15 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum _{n=1}^{\infty }\frac{1+2^n}{3^n}$$

Solution.

The series is a linear combination of two convergent geometric series: $$\sum _{n=1}^{\infty }{\left(\frac{1}{3}\right)}^n\text{and}\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^n.$$ The sums of those two series are $$\sum _{n=1}^{\infty }{\left(\frac{1}{3}\right)}^n=\frac{1}{3}\sum _{n=0}^{\infty }{\left(\frac{1}{3}\right)}^n=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}.$$ and $$\sum _{n=1}^{\infty }{\left(\frac{2}{3}\right)}^n=\frac{2}{3}\sum _{n=0}^{\infty }{\left(\frac{2}{3}\right)}^n=\frac{\frac{2}{3}}{1-\frac{2}{3}}=2.$$ Therefore, $$\sum _{n=1}^{\infty }\frac{1+2^n}{3^n}=\frac{1}{2}+2=\frac{5}{2}.$$

Exercise 10.16 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum _{n=1}^{\infty }(\frac{1}{e^n}+\frac{1}{n(n+1)})$$

Solution.

Consider the series $$\sum _{n=1}^{\infty }{\left(\frac{1}{e}\right)}^n\text{and}\sum _{n=1}^{\infty }\left(\frac{1}{n(n+1)}\right).$$

Because $e>1$, the geometric series $\sum _{n=1}^{\infty }{\left(\frac{1}{e}\right)}^n$ converges to $\frac{1}{e-1}$.

Note that $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. Then the partial sum is $$\begin{array}{cc}{s}_n=& (1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots \\ & +(\frac{1}{n-1}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{n+1})\\ =& 1-\frac{1}{n+1}.\end{array}$$ Taking the limit, we find that $$\sum _{n=1}^{\infty }\left(\frac{1}{n(n+1)}\right)=\underset{n\to \infty }{lim}{s}_n=1.$$

Therefore, $$\sum _{n=1}^{\infty }(\frac{1}{e^n}+\frac{1}{n(n+1)})=\frac{e}{e-1}+1=\frac{2e-1}{e-1}.$$

Exercise 10.17 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum _{n=1}^{\infty }\ln \left(\frac{n}{n+1}\right)$$

Solution.

Then the partial sum is $$\begin{array}{cc}{s}_n=& \sum _{k=1}^n\ln \left(\frac{k}{k+1}\right)\\ =& \sum _{k=1}^n(\ln k-\ln (k+1\left)\right)\\ =& \ln 1-\ln (n+1)=-\ln (n+1).\end{array}$$

Because the limit $-\underset{n\to \infty }{lim}\ln (n+1)=-\infty$. The series in this question diverges.

Exercise 10.18 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum _{n=1}^{\infty }\frac{3}{n(n+3)}$$

Solution.

The rational expression can be written as a sum of partial fraction: $$\frac{3}{n(n+3)}=\frac{1}{n}-\frac{1}{n+3}$$

Then the partial sum is $$\begin{array}{cc}{s}_n=& (1-\frac{1}{4})+(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})\cdots \\ & +(\frac{1}{n-2}-\frac{1}{n+1})+(\frac{1}{n-1}-\frac{1}{n+2})+(\frac{1}{n}-\frac{1}{n+3})\\ =& 1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}.\end{array}$$

Taking the limit, we find that $$\sum _{n=1}^{\infty }\frac{3}{n(n+3)}=\underset{n\to \infty }{lim}{s}_n=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}.$$

Exercise 10.19 Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum _{n=1}^{\infty }(e^{1/n}-e^{1/(n+1)})$$

Solution.

The partial sum is $$\begin{array}{cc}{s}_n=& (e^1-e^{1/2})+(e^{1/2}-e^{1/3})\cdots \\ & +(e^{1/(n-1)}-e^{1/n})+(e^{1/n}-e^{1/(n+1)})\\ =& e-e^{1/(n+1)}.\end{array}$$

Taking the limit, we find that $$\sum _{n=1}^{\infty }\frac{3}{n(n+3)}=\underset{n\to \infty }{lim}{s}_n=e-e^0=e-1.$$

Exercise 10.20 Find the values of $x$ for which the series converges. Find the sum of the series for those values of $x$. $$\sum _{n=1}^{\infty }(x+2{)}^n$$

Solution.

The series is (a multiple of) the geometric series with $r=(x+2)$. Therefore, the series converges if $|x+2|<1$, or equivalently, $-3<x<-1$. For $x\in (-3,-1)$, the sum of the series is $$\sum _{n=1}^{\infty }(x+2{)}^n=(x+2)\cdot \frac{1}{1-(x+2)}=-\frac{x+2}{x+1}.$$

Exercise 10.21 Find the values of $x$ for which the series converges. Find the sum of the series for those values of $x$. $$\sum _{n=0}^{\infty }\frac{2^n}{x^n}$$

Solution.

The series is the geometric series with $r=\frac{2}{x}$. Therefore, the series converges if $|x|>2$, or equivalently, $x\in (-\infty ,-2)\cup (2,\infty )$. When the series converges, the sum of the series is $$\sum _{n=0}^{\infty }{\left(\frac{2}{x}\right)}^n=\frac{1}{1-\frac{2}{x}}=\frac{x}{x-2}.$$

10.3 Integral Test

By comparing the Riemann sum with the (partial) series, we may make the following statement.

Theorem 10.5 (The Integral Test) Suppose $f$ is a continuous, positive, decreasing function on $[1,\infty )$ and let $a_n=f\left(n\right)$. Then the series $\sum _{n=1}^{\infty }{a}_n$ is convergent if and only if the improper integral ${\int }_1^{\infty }f\left(x\right)dx$ is convergent.

As an application, we know exactly when does the $p$-series $\sum \frac{1}{n^p}$ converge.

Corollary 10.1 The $p$-series $\sum _{n=1}^{\infty }\frac{1}{n^p}$ is convergent if $p>1$ and divergent if $p\le 1$.

When a series $\sum a_n$ converges, we may use a partial sum $s_n$ to estimate the sum $s=\sum a_n$. The reminder $R_n=s-s_n$ can be estimated using integral if the function $f$ with $f\left(n\right)=a_n$ is continuous continuous, positive, decreasing function for $x\ge n$ and is convergent: $${\int }_{n+1}^{\infty }f\left(x\right)dx\le R_n\le {\int }_n^{\infty }f\left(x\right)dx$$

Exercise 10.22 Determine whether the series is convergent or divergent. $$1+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}+\frac{1}{4\sqrt{4}}+\frac{1}{5\sqrt{5}}+\cdots$$

Solution.

Rewrite the radicals using rational exponents, we know that the series is a $p$-series $\sum _{n=1}^{\infty }\frac{1}{n^{3/2}}$.

Because $p>1$, the series converges.

Exercise 10.23 Determine whether the series is convergent or divergent. $$\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\frac{1}{17}+\cdots$$

Solution.

The series can be expressed as $$\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}+\frac{1}{17}+\cdots =\sum _{n=1}^{\infty }\frac{1}{2+3n}.$$

Because the improper integral $${\int }_1^{\infty }\frac{1}{3x+2}dx=\frac{1}{3}\left(\underset{t\to \infty }{lim}\ln \right(3x+2)-\ln 5)=\infty ,$$ the series diverges by the Integral Test.

Remark: Can you fill in the intermediate steps for the first equality?
Hint: use substitution $u=3x+2$.

Exercise 10.24 Determine whether the series is convergent or divergent. $$\sum _{n=1}^{\infty }\frac{n^2}{n^3+1}$$

Solution.

Because the improper integral $${\int }_1^{\infty }\frac{x^2}{x^3+1}dx=\frac{1}{3}\left(\underset{t\to \infty }{lim}\ln \right(x^3+1)-\ln 1)=\infty ,$$ the series diverges by the Integral Test.

Remark: Can you fill in the intermediate steps for the first equality?
Hint: use substitution $u=x^3+1$.

Exercise 10.25 Determine whether the series is convergent or divergent. $$\sum _{n=1}^{\infty }\frac{1}{n^2+6n+13}$$

Solution.

Because the improper integral $${\int }_1^{\infty }\frac{1}{x^2+6x+13}dx=\frac{1}{2}(\underset{t\to \infty }{lim}\arctan \left(\frac{t+3}{2}\right)-\arctan 2)=\frac{\pi -2\arctan 2}{4},$$ the series converges by the Integral Test.

Remark: Can you fill in the intermediate steps for the first equality?
Hint: Rewrite the denominator by completing the square $x^2+6x+13=(x+3{)}^2+4$ and use the substitution $x+3=2u$.

Exercise 10.26 Determine whether the series is convergent or divergent. $$\sum _{n=1}^{\infty }\frac{\ln n}{n^3}$$

Solution.

Because the improper integral $$\begin{array}{cc}& {\int }_1^{\infty }\frac{\ln x}{x^3}dx\\ =& (-\underset{t\to \infty }{lim}\frac{\ln t}{2t^2}+\frac{\ln 1}{2})+(-\underset{t\to \infty }{lim}\frac{1}{4t^4}+\frac{1}{4})\\ =& \frac{1}{4},\end{array}$$ the series converges by the Integral Test.

Remark: Can you fill in the intermediate steps for both equalities?
Hint: use integration by part with $g^{\prime }\left(x\right)=\frac{1}{x^3}$ for the first equality and L’Hospital’s rule for the second equality.

Exercise 10.27 Determine whether the series is convergent or divergent. $$\sum _{n=1}^{\infty }\frac{e^{1/n}}{n^2}$$

Solution.

Because the improper integral $$\begin{array}{cc}& {\int }_1^{\infty }\frac{e^{1/x}}{x^2}dx\\ =& -\underset{t\to \infty }{lim}{e}^{1/t}+e^1\\ =& e-1,\end{array}$$ the series converges by the Integral Test.

Remark: Can you fill in the intermediate steps for the first equality?
Hint: use the substitution $u=\frac{1}{x}$.

Exercise 10.28 Find the values $p$ of for which the series is convergent $$\sum _{n=2}^{\infty }\frac{1}{n(\ln n{)}^p}$$

Solution.

Let $u=\ln x$, then $$\int \frac{1}{x(\ln x{)}^p}dx=\{\begin{array}{cc}\ln \left(\ln x\right)& p=1\\ \frac{1}{(1-p)(\ln x{)}^{p-1}}& p\ne 1\end{array}$$

Then

$${\int }_1^{\infty }\frac{1}{x(\ln x{)}^p}dx\{\begin{array}{cc}\infty & p\le 1\\ 0& p>1\end{array}$$ By the Integral Test, the $\sum _{n=2}^{\infty }\frac{1}{n(\ln n{)}^p}$ converges when $p>1$.

Exercise 10.29 Find the sum of the series $\sum _{n=1}^{\infty }\frac{1}{n^3}$ correct to two decimal places.

Solution.

When using the partial sum $s_n$ to approximate the sum, the error (or reminder) $R$ satisfies $$\frac{1}{2(n+1{)}^2}={\int }_{n+1}^{\infty }\frac{1}{x^3}dx\le R\le {\int }_n^{\infty }\frac{1}{x^3}dx=\frac{1}{2n^2}$$

Take $n=8$, we find that $0.006<R<0.008$, so the sum $s\approx s_8\approx 1.195$ is an approximation correct to two decimal places.

10.4 Comparison Test

Comparison is one of fundamental tool that we use to understand new questions. Like sequences, we also have a comparison test of series.

Theorem 10.6 (Comparison Test for Series) Let $\sum a_n$ and $\sum b_n$ be series with positive terms. Suppose $a_n\ge b_n$ for all $n$. If the series $\sum a_n$ converges, then the series $\sum b_n$ converges. Conversely, if the series $\sum b_n$ diverges, then the series $\sum a_n$ diverges.

When compare two numbers, we often compare their quotient with $1$ or other expected number. From this point of view, we may deduce from the comparison test the limit comparison test.

Theorem 10.7 (The Limit Comparison Test for Series) Let $\sum a_n$ and $\sum b_n$ be series with positive terms. Suppose $$\underset{n\to \infty }{lim}\frac{a_n}{b_n}=c,$$ where $c$ is a finite number.
1. If $c$ is positive, then either both the series converge or both diverge.
2. If $c=0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
3. If $c=\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

When using the limit comparison test, it’s better to let $\sum a_n$ to be the given series and take $a_n$ as the numerator in the limit $\underset{n\to \infty }{lim}\frac{a_n}{b_n}$.

Exercise 10.30 Determine whether the series converges or diverges. $$\sum _{n=2}^{\infty }\frac{n^3}{n^4-1}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{\frac{n^3}{n^4-1}}{\frac{1}{n}}=\underset{n\to \infty }{lim}\frac{n^4}{n^4-1}=1$$ and the series $\sum _{n=2}^{\infty }\frac{1}{n}$ diverges, by the limit comparison test, the series $\sum _{n=2}^{\infty }\frac{n^3}{n^4-1}$ diverges.

Remark: The the term is in fraction form, we take the quotient of highest-degree (or largest-exponent) terms in the numerator and denominator as the term $b_n$.

Exercise 10.31 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\frac{n-1}{n^2\sqrt{n}}$$

Solution.

The question of highest degree terms in the fraction is $\frac{n}{n^2\sqrt{n}}=\frac{1}{n^{3/2}}$.

Because $$\underset{n\to \infty }{lim}\frac{\frac{n-1}{n^2\sqrt{n}}}{\frac{1}{n^{3/2}}}=\underset{n\to \infty }{lim}\frac{n-1}{n}=1$$ and the series $\sum _{n=2}^{\infty }\frac{1}{n^{3/2}}$ converges, by the limit comparison test, the series $\sum _{n=2}^{\infty }\frac{n^3}{n^4-1}$ converges.

Exercise 10.32 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\frac{9^n}{3+{10}^n}$$

Solution.

We compare the given series with the geometric series $\sum _{n=1}^{\infty }{\left(\frac{9}{10}\right)}^n$.

Because $$\underset{n\to \infty }{lim}\frac{\frac{9^n}{3+{10}^n}}{\frac{9^n}{{10}^n}}=\underset{n\to \infty }{lim}\frac{3+{10}^n}{{10}^n}=1$$ and the series $\sum _{n=2}^{\infty }{\left(\frac{9}{10}\right)}^n$ converges, by the limit comparison test, the series $\sum _{n=1}^{\infty }\frac{9^n}{3+{10}^n}$ converges.

Exercise 10.33 Determine whether the series converges or diverges. $$\sum _{k=1}^{\infty }\frac{k{\sin }^{2}k}{1+k^3}$$

Solution.

Because $\sum _{k=1}^{\infty }\frac{1}{k^2}$ converges and $$\underset{n\to \infty }{lim}\frac{\frac{k}{1+k^3}}{\frac{1}{k^2}}=1$$ the series $\sum _{k=1}^{\infty }\frac{k}{1+k^3}$ converges.

Because ${\sin }^{2}x\le 1$, by the comparison test, the series $\sum _{k=1}^{\infty }\frac{k{\sin }^{2}k}{1+k^3}$ converges.

Exercise 10.34 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{33n^4+1}}$$

Solution.

Because $\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{3n^4}}$ converges and $$\underset{n\to \infty }{lim}\frac{\frac{1}{\sqrt[3]{33n^4+1}}}{\frac{1}{\sqrt[3]{3n^4}}}=\frac{1}{\sqrt[3]{33}}$$ the series $\sum _{n=1}^{\infty }\frac{1}{\sqrt[3]{33n^4+1}}$ converges.

Exercise 10.35 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\frac{e^{1/n}}{n}$$

Solution.

Because $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges and $$\underset{n\to \infty }{lim}=\frac{\frac{e^{1/n}}{n}}{\frac{1}{n}}=1.$$ The series $$\sum _{n=1}^{\infty }\frac{e^{1/n}}{n}$$ diverges by the limit comparison test.

Exercise 10.36 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\frac{1}{n!}$$

Solution.

Because $\sum _{n=1}^{\infty }\frac{1}{n^2}$ converges. By the limit comparison test, the series $\sum _{n=1}^{\infty }\frac{1}{n(n-1)}$ converges.

Note that $\frac{1}{n!}\le \frac{1}{n(n-1)}$ for $n\ge 2$. Then the series $$\sum _{n=1}^{\infty }\frac{1}{n!}$$ converges by the comparison test.

Exercise 10.37 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\sin \left(\frac{1}{n}\right)$$

Solution.

Because $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges and $$\underset{n\to \infty }{lim}=\frac{\sin \left(\frac{1}{n}\right)}{\frac{1}{n}}=\underset{x\to 0^{+}}{lim}=\frac{\sin x}{x}=1.$$ The series $$\sum _{n=1}^{\infty }\sin \left(\frac{1}{n}\right)$$ diverges by the limit comparison test.

Exercise 10.38 Determine whether the series converges or diverges. $$\sum _{n=2}^{\infty }\frac{7}{\ln n}$$

Solution.

Because $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges and $$\underset{n\to \infty }{lim}=\frac{\frac{7}{\ln n}}{\frac{1}{n}}=\infty .$$ The series $$\sum _{n=1}^{\infty }\frac{1}{\ln n}$$ diverges.

Exercise 10.39 Determine whether the series converges or diverges. $$\sum _{n=1}^{\infty }\frac{\ln n}{n^2}$$

Solution.

Because $\sum _{n=1}^{\infty }\frac{1}{n^{3/2}}$ converges and $$\underset{n\to \infty }{lim}=\frac{\frac{\ln n}{n^2}}{\frac{1}{n^{3/2}}}=\underset{n\to \infty }{lim}\frac{\ln n}{n^{1/2}}=\underset{n\to \infty }{lim}\frac{1}{2n^{1/2}}=0.$$ The series $$\sum _{n=1}^{\infty }\frac{\ln n}{n^2}$$ converges.

Remark: You may find that the limit comparison test is inconclusive if $b_n=\frac{1}{n}$ or $\frac{1}{n^2}$. But taking $b_n=1/n^p$ with $p$ in $(1,2)$ works. Indeed, by the integral test, one can also show the series converges.

10.5 Alternating series

Theorem 10.8 (Alternating Series Test) If the alternating series $$\sum _{n=1}^{\infty }(-1{)}^{n-1}{b}_n=b_1-b_2+b_3-b_4+b_5-b_6+\cdots b_n>0$$ satisfies the following two conditions: (i) $b_{n+1}\le b_n$ for all $n$ and (ii) $\underset{n\to \infty }{lim}{b}_n=0$, then the series converges.

To show that $f\left(n\right)=b_n$ is decreasing, we often show that the sufficient condition $f^{\prime }\left(x\right)<0$ holds.

Exercise 10.40 Determine whether the alternating series converges or diverges. $$\frac{1}{\ln 2}-\frac{1}{\ln 3}+\frac{1}{\ln 4}-\frac{1}{\ln 5}+\frac{1}{\ln 6}-\cdots$$

Solution.

Because $${\left(\frac{1}{\ln x}\right)}^{\prime }=-\frac{1}{x(\ln x{)}^2}<0$$ for $x>2$ and $$\underset{n\to \infty }{lim}\frac{1}{\ln n}=0.$$

Then the given alternating series converges.

Exercise 10.41 Determine whether the alternating series converges or diverges. $$\sum _{n=0}^{\infty }\frac{(-1{)}^{2n+1}}{\sqrt{n+1}}$$

Solution.

Because $(-1{)}^{2n+1}=-1$ for all $n$. The series is not an alternating series. Comparing with the $p$-series $\sum _{n=1}^{\infty }\frac{1}{n^{1/2}}$, we know that $\sum _{n=0}^{\infty }\frac{1}{\sqrt{n+1}}$ diverges. So does $$\sum _{n=0}^{\infty }\frac{(-1{)}^{2n+1}}{\sqrt{n+1}}=-\sum _{n=0}^{\infty }\frac{1}{\sqrt{n+1}}.$$

Exercise 10.42 Determine whether the alternating series converges or diverges. $$\sum _{n=1}^{\infty }(-1{)}^n\frac{\sqrt[3]{3n}}{2n+3}$$

Solution.

Because $$\begin{array}{cc}{\left(\frac{\sqrt[3]{3x}}{2x+3}\right)}^{\prime }=& \frac{\frac{1}{3}{x}^{-2/3}(2x+3)-2x^{1/3}}{(2x+3{)}^2}\\ =& \frac{(3-4x)}{3\sqrt[3]{3x^2}(2x+3{)}^2}<0\end{array}$$ for $x>1$ and $$\underset{n\to \infty }{lim}\frac{\sqrt[3]{3n}}{2n+3}=0.$$

Then the given alternating series converges.

Exercise 10.43 Determine whether the alternating series converges or diverges. $$\sum _{n=1}^{\infty }(-1{)}^{n+1}n{e}^{-n}$$

Solution.

Because $${\left(x{e}^{-x}\right)}^{\prime }=e^{-x}(1-x)<0$$ for $x>2$ and $$\underset{n\to \infty }{lim}n{e}^{-n}=\underset{n\to \infty }{lim}\frac{n}{e^n}=0.$$

Then the given alternating series converges.

Exercise 10.44 Determine whether the alternating series converges or diverges. $$\sum _{n=1}^{\infty }(-1{)}^{n-1}2\arctan n$$

Solution.

Because $\underset{n\to \infty }{lim}2\arctan n=\pi$. The alternating series diverges by the divergence test.

Exercise 10.45 Determine whether the alternating series converges or diverges. $$\sum _{n=1}^{\infty }(-1{)}^n\sin \left(\frac{\pi }{n}\right)$$

Solution.

Because $${\left(\sin \left(\frac{\pi }{x}\right)\right)}^{\prime }=-\frac{\pi \cos \left(\frac{\pi }{x}\right)}{x^2}<0$$ for $x>2$ and $$\underset{n\to \infty }{lim}\sin \left(\frac{\pi }{n}\right)=\underset{x\to 0^{+}}{lim}\sin x=0.$$

The alternating series converges by the alternating convergence test.

Exercise 10.46 Determine whether the alternating series converges or diverges. $$\sum _{n=1}^{\infty }(-1{)}^n\frac{n^n}{n!}$$

Solution.

Because $\frac{n^n}{n!}>1$ for all $n$. Then the divergence test theorem, the series diverges.

Exercise 10.47 Determine whether the alternating series converges or diverges. $$\sum _{n=1}^{\infty }(-1{)}^n(\sqrt{n+1}-\sqrt{n})$$

Solution.

Because $$(\sqrt{x+1}-\sqrt{x}{)}^{\prime }=\frac{\sqrt{x}-\sqrt{x+1}}{2\sqrt{x(x+1)}}<0$$ for $x>0$ and $$\underset{n\to \infty }{lim}(\sqrt{n+1}-\sqrt{n})=\underset{n\to \infty }{lim}\frac{1}{\sqrt{n+1}+\sqrt{n}}=0.$$

The alternating series converges by the alternating convergence test.

10.6 Absolute Convergence and Tests

A series $\sum a_n$ is called absolutely convergent if the series of absolute values $\sum |a_n|$ is convergent.

A series $\sum a_n$ is called conditionally convergent if it is convergent but not absolutely convergent.

By comparing $\sum (a_n+|a_n|)$ with $\sum \left(2|a_n|\right)$, we draw the following conclusion.

Theorem 10.9 If a series $\sum a_n$ is absolutely convergent, then it is convergent.

Inspired by limit comparison test together with the geometric series, we obtain the following tests.

1. Theorem 10.10 (The Ratio Test)
2. If $\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|=L<1$, then the series $\sum _{n=1}^{\infty }{a}_n$ is absolutely convergent.
   
3. If $\underset{n\to \infty }{lim}\left|\frac{a_{n+1}}{a_n}\right|=L>1$, then the series $\sum _{n=1}^{\infty }{a}_n$ is divergent.

1. Theorem 10.11 (The Root Test)
2. If $\underset{n\to \infty }{lim}\sqrt[n]{n\left|a_n\right|}=L<1$, then the series $\sum _{n=1}^n{a}_n$ is absolutely convergent.
   
3. If $\underset{n\to \infty }{lim}\sqrt[n]{n\left|a_n\right|}=L>1$, then the series $\sum _{n=1}^n{a}_n$ is divergent.

The following two facts show that absolutely convergent series behaves better when rearranging terms.

If $\sum a_n$ is an absolutely convergent series with sum $s$, then any rearrangement of $\sum a_n$ has the same sum $s$.

If $\sum a_n$ is a conditionally convergent series and $r$ is any real number, then there is a rearrangement of $\sum a_n$ that has a sum equal to $r$.

Exercise 10.48 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{5n+1}$$

Solution.

The series is an alternating series. Because $\frac{1}{5n+1}$ is decreasing and $\underset{n\to \infty }{lim}\frac{1}{5n+1}=0$. The series is convergent by the alternating convergence test.

Because $$\underset{n\to \infty }{lim}\frac{\frac{1}{5n+1}}{\frac{1}{n}}=\frac{1}{5}$$ and the harmonic series $\sum _{n=0}^{\infty }\frac{1}{n}$ diverges. The absolute series $\sum _{n=0}^{\infty }\frac{1}{5n+1}$ diverges by the limit comparison test.

Exercise 10.49 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=0}^{\infty }\frac{(-3{)}^n}{(2n+1)!}$$

Solution.

Because $$\underset{n\to \infty }{lim}\frac{a_{n+1}}{a_n}=\underset{n\to \infty }{lim}\frac{\frac{3^{n+1}}{\left(2\right(n+1)+1)!}}{\frac{-3^n}{(2n+1)!}}=\underset{n\to \infty }{lim}\frac{3}{(2n+2)(2n+1)}=0.$$ By the ratio test, the series is absolutely convergent.

Exercise 10.50 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=1}^{\infty }\frac{\sin 4n}{4^n}$$

Solution.

Note that $$\frac{|\sin 4n|}{4^n}\le \frac{1}{4^n}.$$

Because the geometric series $$\underset{n\to \infty }{lim}\frac{1}{4^n}$$ converges, by the comparison test, the absolute series $$\underset{n\to \infty }{lim}\frac{|\sin 4n|}{4^n}$$ converges.

Therefore, the series is absolutely convergent.

Exercise 10.51 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=1}^{\infty }\frac{3-2\sin n}{n^{2/3}-2}$$

Solution.

Note that $$1\le 3-2\sin n\le 5$$ and $$\frac{3-2\sin n}{n^{2/3}-2}\ge \frac{1}{n^{2/3}}$$

Because the series $$\sum _{n=1}^{\infty }\frac{1}{n^{2/3}}$$ diverges, by the comparison test, the absolute series $$\sum _{n=1}^{\infty }\frac{3-2\sin n}{n^{2/3}-2}$$ diverges.

Exercise 10.52 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=2}^{\infty }\frac{(-1{)}^n}{\ln n}$$

Solution.

Note that $\ln n<n$ for $n>1$. Then $\frac{1}{\ln n}>\frac{1}{n}$.

Because the series $$\sum _{n=1}^{\infty }\frac{1}{n}$$ diverges, by the comparison test, the absolute series $$\sum _{n=1}^{\infty }\frac{1}{\ln n}$$ diverges.

However, by the alternating convergence test, we know that the series $\sum _{n=2}^{\infty }\frac{(-1{)}^n}{\ln n}$ converges.

Therefore, the series is conditionally convergent.

Exercise 10.53 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=2}^{\infty }{\left(\frac{-2n}{n+1}\right)}^{5n}$$

Solution.

Because the limit $$\sum _{n=1}^{\infty }\sqrt[n]{n{\left(\frac{2n}{n+1}\right)}^{5n}}=\sum _{n=1}^{\infty }{\left(\frac{2n}{n+1}\right)}^5=2^5>1.$$ By the root test, the series diverges.

Exercise 10.54 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\begin{array}{cc}& 1-\frac{1\cdot 3}{3!}+\frac{1\cdot 3\cdot 5}{5!}-\frac{1\cdot 3\cdot 5\cdot 7}{7!}+\cdots \\ & +(-1{)}^{n-1}\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{(2n-1)!}+\cdots \end{array}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n+1)}{(2n+1)!}}{\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (2n-1)}{(2n-1)!}}=\underset{n\to \infty }{lim}\frac{1}{2n}=0.$$ By the ratio test, the series is absolutely convergent.

Exercise 10.55 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum _{n=1}^{\infty }\frac{2^{n^2}}{n!}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{\frac{2^{(n+1{)}^2}}{(n+1)!}}{\frac{2^{n^2}}{n!}}=\underset{n\to \infty }{lim}\frac{2^{2n+1}}{n+1}=\infty .$$ By the ratio test, the series diverges.

10.7 Power series

A power series is a series of the form $$\sum _{n=0}^{\infty }{c}_n{x}^n=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+\cdots ,$$ where $x$ is a variable and $c_n$ are constant numbers called the coefficients of the series.

More general, the series $$\sum _{n=0}^{\infty }{c}_n(x-a{)}^n=c_0+c_1(x-a)+c_2(x-a{)}^2+\cdots$$ is called a power series in $(x-a)$ or a power series centered at $a$ or a power series about $a$.

When a power series converges, we may use it to define a function that make power series very useful.

The interval that consists of all values of for which the series converges is called the interval of convergence of a power series.

By ratio test or root test, we know the following statement is true.

Theorem 10.12 For a given power series $\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$ there are only three possibilities.

1. The series converges only when $x=a$
2. The series converges for all $x$.
3. There is a positive number $R$ such that the series converges if $|x-a|<R$ and diverges if $|x-a|>R$.

The positive number $R$ is called the radius of convergence of the power series. In interval notation, the inequality $|x-a|<R$ can be written as $(a-R,a+R)$. When $x$ is endpoint, that is $x=a-R$ or $x=a+R$, the series might converge or diverge. So to find the interval of convergence, the convergence of the series $\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$ at endpoints $x=a\pm R$ should be inspected carefully.

Exercise 10.56 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=1}^{\infty }(-1{)}^{n}n{x}^n$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{(n+1)|x{|}^{n+1}}{n|x{|}^n}=|x|.$$

By ratio test, we know the the series converges if $|x|<1$. So the radius of convergence is $R=1$.

When $|x|=1$, the limit $\underset{n\to \infty }{lim}n=\infty$ so the power series diverges when $|x|=1$.

Therefore, the interval of convergence is $(-1,1)$.

Exercise 10.57 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^n}{n^2}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{n^2|x{|}^{n+1}}{(n+1{)}^2|x{|}^n}=|x|.$$

By ratio test, we know the the series converges if $|x|<1$. So the radius of convergence is $R=1$.

When $|x|=1$, the absolute value series is the $p$-series with $p=2>1$. It follows that the power series $\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^n}{n^2}$ is absolutely convergent when $|x|=1$.

Therefore, the interval of convergence is $[-1,1]$.

Exercise 10.58 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=1}^{\infty }(-1{)}^n\frac{n^2{x}^n}{2^n}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{(n+1{)}^2{2}^n|x{|}^{n+1}}{n^2{2}^{n+1}|x{|}^n}=\frac{|x|}{2}.$$

By ratio test, we know the the series converges if $\frac{|x|}{2}<1$, equivalently, $|x|<2$. So the radius of convergence is $R=2$.

When $|x|=2$, ${lim}_{n\to \infty }\frac{(-1{)}^n{n}^2{x}^n}{2^n}$ does not exist. By the divergence test, the power series diverges when $|x|=2$.

Therefore, the interval of convergence is $(-2,2)$.

Exercise 10.59 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{\frac{|x{|}^{2n+3}}{(2n+3)!}}{\frac{|x{|}^{2n+1}}{(2n+1)!}}=\underset{n\to \infty }{lim}\frac{|x{|}^2}{(2n+3)(2n+2)}=0,$$ for any $x$.

By ratio test, we know the the series converges for all $x$. So the radius of convergence is $R=\infty$ and the interval of convergence is $(-\infty ,\infty )$.

Exercise 10.60 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=1}^{\infty }\frac{3^n(x+4{)}^n}{\sqrt{n}}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{\frac{3^{n+1}|x+4{|}^{n+1}}{\sqrt{n+1}}}{\frac{3^n|x+4{|}^n}{\sqrt{n}}}=\underset{n\to \infty }{lim}\frac{3|x+4|\sqrt{n}}{\sqrt{n+1}}=3|x+4|.$$

By ratio test, we know the the series converges if $|x+4|<\frac{1}{3}$. So the radius of convergence is $R=\frac{1}{3}$.

When $x+4=\frac{1}{3}$, the power series is a divergent $p$-series with $p=\frac{1}{2}$.

When $x+4=-\frac{1}{3}$, the power series is a convergent alternating series.

Therefore, the interval of convergence is $$[-\frac{1}{3}-4,\frac{1}{3}-4)=[-\frac{13}{3},-\frac{11}{3}).$$

Exercise 10.61 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=2}^{\infty }\frac{b^n}{\ln n}(x-a{)}^n,b>0$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{\frac{b^{n+1}|x-a{|}^{n+1}}{\ln (n+1)}}{\frac{b^n|x-a{|}^n}{\ln n}}=\underset{n\to \infty }{lim}\frac{b|x-a|\ln n}{\ln (n+1)}=b|x-a|.$$

By ratio test, we know the the series converges if $|x-a|<\frac{1}{b}$. So the radius of convergence is $R=\frac{1}{b}$.

When $x-a=\frac{1}{b}$, the power series is a divergent by comparing with the $p$-series $\sum _{n\to \infty }\frac{1}{n}$.

When $x-a=-\frac{1}{b}$, the power series is a convergent alternating series.

Therefore, the interval of convergence is $$[-\frac{1}{b}+a,\frac{1}{b}+a)=[\frac{ab-1}{b},-\frac{ab+1}{b}).$$

Exercise 10.62 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=1}^{\infty }n!(2x-1{)}^n$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{(n+1)!|2x-1{|}^{n+1}}{n!|2x-1{|}^n}=\{\begin{array}{cc}0& \text{if}x=\frac{1}{2}\\ \infty & \text{otherwise}\end{array}.$$

By ratio test, we know the the series converges only if $x=\frac{1}{2}$. So the radius of convergence is $R=0$.

Therefore, the interval of convergence is a one-point set $\left\{\frac{1}{2}\right\}$.

Exercise 10.63 Find the radius of convergence and interval of convergence of the series. $$\sum _{n=2}^{\infty }\frac{x^{2n}}{n(\ln n{)}^2}$$

Solution.

Because the limit $$\underset{n\to \infty }{lim}\frac{|a_{n+1}|}{|a_n|}=\underset{n\to \infty }{lim}\frac{\frac{|x{|}^{2n+2}}{(n+1)\left(\ln \right(n+1){)}^2}}{\frac{|x{|}^{2n}}{n(\ln n{)}^2}}=\underset{n\to \infty }{lim}\frac{n(\ln n{)}^2|x{|}^2}{(n+1)\left(\ln \right(n+1){)}^2}=|x{|}^2.$$

By ratio test, we know the the series converges if $|x{|}^2<1$, equivalently, $|x|<1$. So the radius of convergence is $R=1$.

When $|x|=1$, the power series is a convergent by the integral test.

Therefore, the interval of convergence is $[-1,1]$.

10.8 Representations of Functions as Power Series

Given a power series $\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$, we may define a function $f\left(x\right)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$ over the interval of convergence.

For example, $$\frac{1}{1-x}=1+x+x^2+x^3+\cdots =\sum _{n=0}^{\infty }{x}^n\text{for}|x|<1.$$

Over the open interval $(a-R,a+R)$, where $R$ is the radius of convergence, differential and integration commute with the sum operation.

Theorem 10.13 If the power series $\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$ has radius of convergence $R$, then the function $f$ defined by $$f\left(x\right)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n$$ is differentiable over the interval $(a-R,a+R)$. Moreover, $$\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left[\sum _{n=0}^{\infty }{c}_n\right(x-a{)}^n]=\sum _{n=0}^{\infty }\frac{d}{dx}\left[c_n\right(x-a{)}^n]$$ and $$\int f\left(x\right)dx=\int \left[\sum _{n=0}^{\infty }{c}_n\right(x-a{)}^n]dx=\sum _{n=0}^{\infty }\int \left[c_n\right(x-a{)}^{n}dx]$$

Exercise 10.64 Find a power series representation for the function and determine the radius of convergence. $$f\left(x\right)=\frac{1}{2-x}$$

Solution.

Using the power series representation $$\frac{1}{1-x}=1+x+x^2+\cdots ,$$

We obtain $$\begin{array}{cc}\frac{1}{2-x}=& \frac{1}{2}\cdot \frac{1}{1-\frac{x}{2}}\\ =& \frac{1}{2}\cdot (1+\frac{x}{2}+{\left(\frac{x}{2}\right)}^2+\cdots )\\ =& \frac{1}{2}+\frac{x}{2^2}+\frac{x^2}{2^3}+\cdots \\ =& \sum _{n=0}^{\infty }\frac{x^n}{2^{n+1}}\end{array}$$

By ratio test, we find the radius of convergence is $R=2$.

Exercise 10.65 Find a power series representation for the function and determine the radius of convergence. $$f\left(x\right)=\frac{x}{4+x^2}$$

Solution.

A power series expansion can be obtained as follows $$\begin{array}{cc}\frac{x}{4+x^2}=& \frac{x}{4}\cdot \frac{1}{1-(-\frac{x^2}{4})}\\ =& \frac{x}{4}\cdot (1+(-\frac{x^2}{4})+{(-\frac{x^2}{4})}^2+\cdots )\\ =& \frac{x}{4}-\frac{x^3}{4^2}+\frac{x^5}{4^3}-\cdots \\ =& \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{4^{n+1}}.\end{array}$$

By ratio test, we find the radius of convergence is $R=2$. Because the power series converges when $|x{|}^2/4<1$.

Exercise 10.66 Find a power series representation for the function and determine the radius of convergence. $$f\left(x\right)=\ln (5-x)$$

Solution.

Similar to the first exercise, we have a power series representation for $\frac{1}{5-x}$

$$\frac{1}{5-x}=\sum _{n=0}^{\infty }\frac{x^n}{5^{n+1}}.$$

Because $\left(\ln \right(5-x){)}^{\prime }=-\frac{1}{5-x}$. By taking integral, we have $$\begin{array}{cc}\ln (5-x)=& \int -\frac{1}{5-x}dx\\ =& \int \left(\sum _{n=0}^{\infty }\frac{x^n}{5^{n+1}}\right)dx\\ =& \sum _{n=0}^{\infty }\int \frac{x^n}{5^{n+1}}dx\\ =& -\sum _{n=0}^{\infty }\frac{x^{n+1}}{(n+1)5^{n+1}}+C\\ =& -\sum _{n=0}^{\infty }\frac{x^{n+1}}{(n+1)5^{n+1}}+\ln 5.\end{array}$$ The last step was obtained by plugging $x=0$ into both sides.

The radius of convergent is $R=5$ by a ratio test.

Exercise 10.67 Find a power series representation for the function and determine the radius of convergence. $$f\left(x\right)=x^2\arctan \left(x^3\right)$$

Solution.

If we can find a power series representation for $\arctan \left(x^3\right)$, then multiplying the power series by $x^2$, we will get a power series representation for $x^2\arctan \left(x^3\right)$.

Note that $$\begin{array}{cc}\left(\arctan \right(x^3){)}^{\prime }=& \frac{3x^2}{1+x^6}\\ =& 3x^2\left(\frac{1}{1-(-x^6)}\right)\\ =& 3x^2(1+(-x^6)+(-x^6{)}^2+(-x^6{)}^3+\cdots )\\ =& \sum _{n=0}^{\infty }3(-1{)}^n{x}^{6n+2}.\end{array}$$

Then $$\begin{array}{cc}\arctan \left(x^3\right)=& \sum _{n=0}^{\infty }\frac{3(-1{)}^n{x}^{6n+3}}{6n+3}+C\\ =& \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{6n+3}}{2n+1}\text{because}\arctan 0=0\end{array}$$

Therefore, $$x^2\arctan \left(x^3\right)\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{6n+5}}{2n+1}.$$

The radius of convergent is $R=1$ by a ratio test.

Exercise 10.68 Find a power series representation for the function and determine the radius of convergence. $$f\left(x\right)={\left(\frac{x}{2-x}\right)}^3$$

Solution.

A power series representation can be obtained as follows $$\begin{array}{cc}{\left(\frac{x}{2-x}\right)}^3=& x^3{\left(\frac{1}{2-x}\right)}^3\\ =& x^3\cdot \frac{1}{2}\cdot {\left(\frac{1}{2-x}\right)}^{\prime \prime }\\ =& \frac{1}{2}{x}^3{\left(\sum _{n=0}^{\infty }\frac{x^n}{2^{n+1}}\right)}^{\prime \prime }\\ =& \frac{1}{2}{x}^3\left(\sum _{n=2}^{\infty }\frac{n(n-1)x^{n-2}}{2^{n+1}}\right)\\ =& \sum _{n=2}^{\infty }\frac{n(n-1)x^{n+1}}{2^{n+2}}.\end{array}$$

The radius of convergent is $R=2$ by a ratio test.

Exercise 10.69 Evaluate the indefinite integral as a power series. Find the radius of convergence of the integral. $$\int \frac{t}{1-t^8}dt$$

Solution.

The rational function $f\left(t\right)=\frac{t}{1-t^8}$ has a power series representation $$\frac{t}{1-t^8}=t\cdot \frac{1}{1-t^8}=t\sum _{n=0}^{\infty }{t}^{8n}=\sum _{n=0}^{\infty }{t}^{8n+1}.$$

Therefore, the integral has a power series representation as follows $$\begin{array}{cc}\int \frac{t}{1-t^8}dt=& \int \left(\sum _{n=0}^{\infty }{t}^{8n+1}\right)dt\\ =& \sum _{n=0}^{\infty }(\int t^{8n+1}dt)\\ =& \sum _{n=0}^{\infty }\frac{t^{8n+2}}{8n+2}+C\end{array}$$

The radius of convergent is $R=1$ by a ratio test.

Exercise 10.70 Evaluate the indefinite integral as a power series. Find the radius of convergence of the integral $$\int \frac{\arctan x}{x}dx$$

Solution.

The function $f\left(x\right)=\arctan x$ has a power series representation as follows $$\begin{array}{cc}\arctan x=& \int \frac{1}{1+x^2}dx+C_0\\ =& \int \sum _{n=0}^{\infty }(-1{)}^n{x}^{2n}dx+C_0\\ =& \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1}+C\\ =& \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1}\end{array}$$

Therefore, the integral has a power series representation $$\begin{array}{cc}\int \frac{\arctan x}{x}dx=& \int \left(\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{2n+1}\right)dt\\ =& \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1{)}^2}+C\end{array}$$

The radius of convergent is $R=1$ by a ratio test.

Exercise 10.71 Use a power series to approximate the definite integral to six decimal places. $${\int }_0^{0.2}\frac{1}{1+x^5}dx$$

Solution.

The indefinite integral has a power series representation as follows $$\begin{array}{cc}\int \frac{1}{1+x^5}dx=& \int \sum _{n=0}^{\infty }(-1{)}^n{x}^{5n}dx\\ =& \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{5n+1}}{5n+1}+C\end{array}$$

The radius is $R=1$.

Since $\frac{{0.2}^{11}}{11}\approx 1.89\times {10}^{-9}$, to approximate the integral to six decimal places, it is enough to use only the first two terms of the power series representation, that is $${\int }_0^{0.2}\frac{1}{1+x^5}dx\approx \frac{0.2}{1}-\frac{{0.2}^6}{6}\approx 0.2-0.000011=\mathrm{0.199989.}$$

10.9 Taylor Series and Applications

Suppose a function $f$ has a power series representation, by differentiating the representation, we find the coefficients of the power series representation are determined by derivatives of the function.

Theorem 10.14 (Taylor Coefficients) If the function $f$ has a power series representation (expansion) at $a$, that is, $$f\left(x\right)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n|x-a|<R,$$ then the coefficients $c_n$ are determined by the formula $$c_n=\frac{f^{\left(n\right)}\left(a\right)}{n!}.$$ In other words, if $f$ has power series expansion, then it must be $$f\left(x\right)=f\left(a\right)+\frac{f^{\prime }\left(a\right)}{1!}(x-a)+\frac{f^{\prime \prime }\left(a\right)}{2!}(x-a{)}^2+\frac{f^{\prime \prime \prime }\left(a\right)}{3!}(x-a{)}^3+\cdots .$$

For an infinitely differentiable function $f$, we may define a power series at $a$ $$\sum _{n=0}^{\infty }\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k.$$ We call this series the Taylor series generated by $f$ at $a$ or the Taylor expansion of $f$ at $a$.

The Taylor series of $f$ at $0$ is also called an Maclaurin series (expansion).

In general, the Taylor series may not converge for $x\ne a$. Even it converges for $x\ne a$, the sum of the Taylor series may not be $f\left(x\right)$.

Remark. A function $f$ that has a convergent power series representation is called a (real) analytic function. Real analytical functions are infinitely differentiable (also known as smooth), that is, differentiable up to any order.

However, not all smooth functions are real analytic. There exists non-analytic smooth functions.

A sufficient (and necessary) condition for the convergence of the Taylor series to $f\left(x\right)$ for $x\ne a$ can be given using the reminder of the Taylor series.

We call the partial sum $$T_n\left(x\right)=\sum _{k=0}^n\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k$$ the $n$-th degree Taylor polynomial of $f$ at $a$.

If $f$ is the sum of its Taylor series, then $\underset{n\to \infty }{lim}{T}_n\left(x\right)=f\left(x\right)$ or equivalently $\underset{n\to \infty }{lim}\left(f\right(x)-T_n(x\left)\right)=0$. We call the difference $R_n\left(x\right)=f\left(x\right)-T_n\left(x\right)$ the reminder of the Taylor series.

Theorem 10.15 (Taylor Formula) Assume that $f$ has a continuous derivative of order $n+1$ in an interval $I$ of $a$. Then for any $x$ in $I$, we have $$f\left(x\right)=\sum _{k=0}^n\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k+R_n(x),$$ where $$R_n\left(x\right)=\frac{1}{n!}{\int }_a^x(x-t{)}^n{f}^{(n+1)}(t)dt=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}(x-a{)}^{n+1},$$ and $c$ is in between $a$ and $x$ inclusively.

The Taylor formula and the first equality for $R_n\left(x\right)$ can be proved by induction. The second equality for $R_n\left(x\right)$ is obtained by the weighted mean-value theorem for integral. The reminder $R_n\left(x\right)=\frac{f^{(n+1)}\left(c\right)}{(n+1)!}(x-a{)}^{n+1}$ is called the Lagrange’s form of the reminder.

From Theorem 10.15, we can deduce a sufficient condition such that the sum of the Taylor series of $f$ equals $f\left(x\right)$.

Theorem 10.16 If $|f^{(n+1)}\left(x\right)|\le M$ for $|x-a|\le R$ and $$\underset{n\to \infty }{lim}\frac{M}{(n+1)!}|x-a{|}^{n+1}=0\text{for}|x-a|\le R.$$ Then $$f\left(x\right)=\sum _{n=0}^{\infty }{c}_n(x-a{)}^n\text{for}|x-a|<R.$$

One can check that all elementary functions are sums of their Taylor series.

Important Maclaurin Series and their Radii of Convergence

Power Series Representation	Radius of Convergence	Interval of Convergence
$\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^n=1+x+x^2+x^3+\cdots$	$R=1$	$(-1,1)$
$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$	$R=\infty$	$(-\infty ,\infty )$
$\sin x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$	$R=\infty$	$(-\infty ,\infty )$
$\cos x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{\left(2n\right)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$	$R=\infty$	$(-\infty ,\infty )$
$\arctan x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{2n+1}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$	$R=1$	$[-1,1]$
$\ln (1+x)=\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}{x}^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$	$R=1$	$(-1,1]$
$(1+x{)}^k=\sum _{n=0}^{\infty }(\frac{k}{n})x^n=1+kx+\frac{k(k-1)}{2!}{x}^2+\frac{k(k-1)(k-2)}{3!}{x}^3+\cdots$	$R=1$	depends on $k$

Remark. Using Taylor expansion, we can deduce the famous Euler’s formula $$e^{i\theta }=\cos \theta +i\sin \theta$$ for all real numbers $\theta$.

Exercise 10.72 Find the Taylor series for the given function at the given point. Determine the radius of convergence. $$f\left(x\right)=\ln x,a=1.$$

Solution.

By induction, the $n$-th derivative of $f\left(x\right)=\ln x$ is $$f^{\left(n\right)}\left(x\right)=\frac{d^{n-1}}{d{x}^{n-1}}\left(x^{-1}\right)=(-1{)}^{n+1}(n-1)!x^{-n}.$$ Plugging in $x=1$, we obtain the $n$-th Taylor coefficient $$\frac{f^{\left(n\right)}\left(1\right)}{n!}=\frac{(-1{)}^{n+1}(n-1)!}{n!}=\frac{(-1{)}^{n+1}}{n}.$$

Therefore, the Taylor series of the function $f\left(x\right)=\ln x$ at $a=1$ is $$\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}(x-1)}{n}=(x-1)-\frac{(x-1{)}^2}{2}+\frac{(x-1{)}^3}{3}-\frac{(x-1{)}^4}{4}+\cdots .$$ By the ratio test, we know the radius of convergence of the Taylor series is $R=1$.

Remark: It can be shown that $\ln x=\sum _{n=1}^{\infty }\frac{(-1{)}^{n+1}(x-1)}{n}$ using the power series representation $\ln (1+x)=\sum _{n=0}^{\infty }\frac{x^n}{n}$.

Exercise 10.73 Find the Maclaurin series for the given function at the given point. Determine the radius of convergence. $$f\left(x\right)=\arctan \left(x^2\right)$$

Solution.

Using the power series expansion $$\arctan x=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{2n+1}$$ we get $$\arctan \left(x^2\right)=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{4n+2}}{2n+1}.$$ By Theorem 10.14, we know that $\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{4n+2}}{2n+1}$ is the Maclaurin series of $f$.

The radius of convergence is $R=1$.

Exercise 10.74 Find the Maclaurin series for the given function at the given point. Determine the radius of convergence. $$f\left(x\right)=x\ln (1+x^2)$$

Solution.

Using the power series expansion $$\ln (1+x)=\sum _{n=0}^{\infty }\frac{x^n}{n}$$ we get $$x\ln (1+x^2)=x\sum _{n=0}^{\infty }\frac{x^{2n}}{n}=\sum _{n=0}^{\infty }\frac{x^{2n+1}}{n}$$ which is the Maclaurin series of $f\left(x\right)=x\ln (1+x^2)$.

The radius of convergence is $R=1$.

Exercise 10.75 Find the Maclaurin series for the given function at the given point. Determine the radius of convergence. $$f\left(x\right)=\sin \left(\pi x\right)$$

Solution.

Using the power series expansion $$\sin x=\sum _{n=0}^{\infty }(-1{)}^n\frac{x^{2n+1}}{(2n+1)!}$$ we get the Maclaurin series $$\sin \left(\pi x\right)=\sum _{n=0}^{\infty }(-1{)}^n\frac{{\pi }^{2n+1}{x}^{2n+1}}{(2n+1)!}$$

The radius of convergence is $R=\infty$.

Exercise 10.76 Find the Maclaurin series for the given function at the given point. Determine the radius of convergence. $$f\left(x\right)=x^2{e}^{2x}$$

Solution.

Using the power series expansion $$e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$ we get the Maclaurin series $$x^2{e}^{2x}=x^2\sum _{n=0}^{\infty }\frac{2^n{x}^n}{n!}=\sum _{n=0}^{\infty }\frac{2^n{x}^{n+2}}{n!}.$$

The radius of convergence is $R=\infty$.

Exercise 10.77 Find the Maclaurin series for the given function at the given point. Determine the radius of convergence. $$f\left(x\right)=\frac{x}{\sqrt{4+x}}$$

Solution.

Using the power series expansion $$(1+x{)}^k=1+kx+\frac{k(k-1)}{2!}{x}^2+\frac{k(k-1)(k-2)}{3!}{x}^3+\cdots$$ we get $$(1+x{)}^{-1/2}=1-\frac{1}{2}x+\frac{1\cdot 3}{2^2\cdot 2!}{x}^2-\frac{1\cdot 3\cdot 5}{2^3\cdot 3!}{x}^3+\cdots =\sum _{n=0}^{\infty }(-1{)}^n\frac{(2n-1)!}{2^{2n-1}(n-1)!n!}{x}^n$$ Then the Maclaurin series of $\frac{x}{\sqrt{4+x}}$ is $$\begin{array}{cc}\frac{x}{\sqrt{4+x}}=& \frac{x}{2}\cdot (1+\frac{x}{2}{)}^{-1/2}\\ =& \frac{x}{2}(1-\frac{1}{2}\left(\frac{x}{2}\right)+\frac{1\cdot 3}{2^2\cdot 2!}{\left(\frac{x}{2}\right)}^2-\frac{1\cdot 3\cdot 5}{2^3\cdot 3!}{\left(\frac{x}{2}\right)}^3+\cdots )\\ =& \frac{x}{2}(1-\frac{1}{2\cdot 2}x+\frac{1\cdot 3}{2^4\cdot 2!}{x}^2-\frac{1\cdot 3\cdot 5}{2^6\cdot 3!}{x}^3+\cdots )\\ =& \frac{x}{2}-\frac{1}{2^3}{x}^2+\frac{1\cdot 3}{2^5\cdot 2!}{x}^3-\frac{1\cdot 3\cdot 5}{2^7\cdot 3!}{x}^4+\cdots \\ =& \sum _{n=0}^{\infty }(-1{)}^n\frac{(2n-1)!}{2^{3n}(n-1)!n!}{x}^{n+1}.\end{array}$$

Exercise 10.78 Evaluate the integral as a power series and determine the radius of convergence. $$\int \frac{\cos x-1}{x}dx$$

Solution.

Because $$\cos x=\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{\left(2n\right)!}$$ with the radius of convergence $R=\infty$. Then $$\frac{\cos x-1}{x}=\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{2n-1}}{\left(2n\right)!}.$$ with the radius of convergence $R=\infty$.

Taking integral of both sides, we get $$\begin{array}{cc}\int \frac{\cos x-1}{x}dx=& \int \left(\sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{2n-1}}{\left(2n\right)!}\right)dx\\ =& \sum _{n=1}^{\infty }(\int \frac{(-1{)}^n{x}^{2n-1}}{\left(2n\right)!}dx)\\ =& \sum _{n=1}^{\infty }\frac{(-1{)}^n{x}^{2n}}{2n\left(2n\right)!}.\end{array}$$

Exercise 10.79 Evaluate the limit $$\underset{x\to 0}{lim}\frac{x-\arctan x}{x^3}$$

Solution.

Using the Taylor expansion $$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}$$ we find that $$\frac{x-\arctan x}{x^3}=-\frac{1}{3}+\frac{x^2}{5}-\frac{x^4}{7}+\cdots$$ Using the fact that $-\frac{1}{3}+\frac{x^2}{5}-\frac{x^4}{7}+\cdots$ is a continuous function, we find that $$\begin{array}{cc}\underset{x\to 0}{lim}\frac{x-\arctan x}{x^3}=& \underset{x\to 0}{lim}(-\frac{1}{3}+\frac{x^2}{5}-\frac{x^4}{7}+\cdots )\\ =& -\frac{1}{3}+\frac{0^2}{5}-\frac{0^4}{7}+\cdots \\ =& -\frac{1}{3}.\end{array}$$

Exercise 10.80 Find the sum of the series $$\frac{\pi }{1!\cdot 2}-\frac{{\pi }^3}{3!\cdot 2^3}+\frac{{\pi }^5}{5!\cdot 2^5}-\frac{{\pi }^7}{7!\cdot 2^7}+\cdots$$

Solution.

Consider the power series $$\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ which converges for all real number $x$.

Comparing with the Taylor expansion of $\sin x$, we find that $$\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sin x$$ for all real number $x$.

Plug $x=\frac{\pi }{2}$ into both sides, we get $$\frac{\pi }{1!\cdot 2}-\frac{{\pi }^3}{3!\cdot 2^3}+\frac{{\pi }^5}{5!\cdot 2^5}-\frac{{\pi }^7}{7!\cdot 2^7}+\cdots =\sin \frac{\pi }{2}=1.$$

Exercise 10.81 Find the sum of the series $$\frac{1}{2}+\frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}+\cdots$$

Solution.

Consider the power series $$x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots$$ which converges when $|x|<1$.

Comparing with the Taylor expansion of $\ln (1+x)$, we find that $$-\ln (1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots$$ for $|x|<1$.

Plug $x=\frac{1}{2}$ into both sides, we get $$\frac{1}{2}+\frac{1}{2\cdot 2^2}+\frac{1}{3\cdot 2^3}+\frac{1}{4\cdot 2^4}+\cdots =-\ln (1-\frac{1}{2})=\ln 2.$$