Topic 10 Sequences and Series

10.1 Sequences

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

The limit limn→∞an of a sequence {an} is defined similarly to the limit limn→∞f(x) of the function f. Indeed, we have the following theorem.

Theorem 10.1 If limx→∞f(x)=L and f(n)=an, then limn→∞an=L.

If the limit limn→∞an exists, we say the sequence {an} is convergent. Otherwise, we say the sequence {an} is divergent.

Convergence sequences have the same rules and properties of limits as functions. For example, we may apply L’Hospital’s rule to 00 or ∞∞ 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 {an} be a bounded sequence, that is |an|<M for all n. If there exists an integer N such that either an≤an+ for all n>N or an≥an+ 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 an=2nn2+1

Solution.

The first five terms are 1,45,35,817,513

Exercise 10.2 List the first five terms of the sequence a1=2,a2=1,an+1=an−an−1

Solution.

The first two terms have been given. Using the recurrence formula, we find a3=−1,a4=−2,a5=−1.

Exercise 10.3 Find a formula for the general term an of the following sequence, assuming that the pattern of the first few terms continues. {12,−43,94,−165,256,…}

Solution.

The sequence can be explicitly expressed as {(−1)n+1n2n+1}∞n=1

Exercise 10.4 Determine whether the sequence converges or diverges. If it converges, find the limit. an=3+5n2n+n2

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. limn→∞3+5n2n+n2=limn→∞3n2+51n+1=limn→∞(3n2)+5limn→∞(1n)+1=5. The sequence converges to 5.

Exercise 10.5 Determine whether the sequence converges or diverges. If it converges, find the limit. an=3n+25n

Solution.

We rewrite the expression in the form a⋅rn form and use the fact that limn→∞rn=0 if |r|<1 to determine the convergence.

limn→∞3n+25n=9limn→∞(35)n=9⋅0=0. The sequence converges to 0.

Exercise 10.6 Determine whether the sequence converges or diverges. If it converges, find the limit. an=e2n/(n+2)

Solution.

Let f(x)=e2x/(x+2). Then f is continuous whenever x>0. We may use the theorem that limn→∞e2n/(n+2)=limx→∞e2x/(x+2) to determine the convergence.

Use the fact that y=ex is continuous and limx→∞(2x/(x+2))=2, we get limx→∞e2x/(x+2)=elimx→∞(2x/(x+2))=e2. The sequence converges to e2.

Exercise 10.7 Determine whether the sequence converges or diverges. If it converges, find the limit. {(2n−1)!(2n+1)!}

Solution.

Recall that n!=n⋅(n−1)⋅(n−2)⋅⋯⋅2⋅1.

Then (2n−1)!(2n+1)!=(2n−1)⋅(2n−2)⋅⋯⋅2⋅1(2n+1)⋅2n⋅(2n−1)⋅(2n−2)⋅⋯⋅2⋅1=1(2n+1)⋅2n Therefore, limn→∞{(2n−1)!(2n+1)!}=limx→∞1(2n+1)⋅2n=0.

The sequence converges to 0.

Exercise 10.8 Determine whether the sequence converges or diverges. If it converges, find the limit. {n2e−n}

Solution.

Let f(x)=x2e−x. Then f is continuous and limx→∞n2e−n=limx→∞x2e−x=limx→∞x2ex=limx→∞2xexL'Hospital's Rule Applied=limx→∞2exL'Hospital's Rule Applied=0. The sequence converges to 0.

Exercise 10.9 Determine whether the sequence converges or diverges. If it converges, find the limit. an=nsin(1/n)

Solution.

Let f(x)=xsin(1/x). Then f is continuous whenever x>0. Therefore, limx→∞nsin(1/n)=limx→∞xsin(1/x)=limx→∞sin(1/x)1/x=limt→0+sinttt=1/x=limt→0+cost1L'Hospital's Rule Applied=1. The sequence converges to 1.

Exercise 10.10 Determine whether the sequence converges or diverges. If it converges, find the limit. a1=3 and an+1=an2+1 for all n≥2

Solution.

By the recurrence formula, we get a1=3,a2=52,a3=94,a4=178.

From those terms, we may claim that the sequence is decreasing and an≥2.

We prove the claim that an≥2 by induction. The monotonicity follows from this claim.

Clearly, a1≥2. Suppose that the an>2. Then an+1=an2+1≥1+1=2.

By the Monotone Convergence Theorem, limn→∞an=L is a finite number.

Taking limits of both side of the equality an+1=an2+1, we get L=L2+1. Solve the equation for L, we get limn→∞an=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=n0, followed by a inductive step. In the inductive step, we assume the claim is true for k=n (or k≤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. a1=1an+1=3−1an

Solution.

By the recurrence formula, we get a1=1,a2=2,a3=3−12=52,a4=3−25=135.

From those terms, we may claim that the sequence is increasing and 1≤an≤3.

To prove that the sequence is increasing, we need to show that an+1−an=3−1an−an=−a2n+3an−1an>0. If we can prove that an≤3+√52, then an+1−an>0. Because −a2n+3an−1>0 when 1≤an≤3+√52.

This can be done by induction.

Clearly, 1≤a1≤3+√52 Suppose that the 1≤an≤3+√52. Then 1<an+1=3−1an≤3−23+√5=3+√52.

By the Monotone Convergence Theorem, limn→∞an=L is a finite number.

Taking limits of both side of the equality an+1=3−1an, we get L=3−1L. Solve the equation, we get limn→∞an=L=3+√52.

10.2 Series

A series ∞∑n=1an is the (formal) sum of all terms of a sequence {an}∞n=1. The sum sn=a1+a2+a3+⋯+an=n∑i=1ai is called a partial sum of the series.

A series ∞∑n=1an is convergent (divergent) if the sequence of partial sums {sn} is convergent (respectively, divergent).

Theorem 10.3 If a series ∞∑n=1an is convergent then limn→∞an=0. Conversely, if limn→∞an≠0, then ∞∑n=1an is divergent.

For convergence series, their linear combinations are also convergent.

Theorem 10.4 If ∞∑n=1an and ∞∑n=1bn are convergence sequences, then ∞∑n=1(c⋅an+d⋅bn)=c∞∑n=1an+d∞∑n=1bn, where c and d are real numbers.

Using the formula 1+r+r2+r3+⋯+rn−1=1−rn1−r we know that the geometric series ∞∑n=0a⋅rn=a1−rif|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+r2+r3+⋯) 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+94+2716+⋯

Solution.

The series can be expressed as 4⋅(1+34+(34)2+(34)3+⋯) Which is a scalar multiple of the geometric series with r=34<1.

Therefore, the series converges and 4+3+94+2716+⋯=41−34=16.

Exercise 10.13 Determine whether the series is convergent or divergent. If it is convergent, find its sum. ∞∑n=1(−3)n−14n

Solution.

The series can be expressed as 13⋅(1+(−34)+(−34)2+(−34)3+⋯).

Therefore, the series converges and ∞∑n=1(−3)n−14n=141−(−34)=17.

Exercise 10.14 Determine whether the series is convergent or divergent. If it is convergent, find its sum. 13+16+19+112+115+⋯

Solution.

The series can be expressed as 14⋅(1+12+13+14+15+⋯). 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. ∞∑n=11+2n3n

Solution.

The series is a linear combination of two convergent geometric series: ∞∑n=1(13)nand∞∑n=1(23)n. The sums of those two series are ∞∑n=1(13)n=13∞∑n=0(13)n=131−13=12. and ∞∑n=1(23)n=23∞∑n=0(23)n=231−23=2. Therefore, ∞∑n=11+2n3n=12+2=52.

Exercise 10.16 Determine whether the series is convergent or divergent. If it is convergent, find its sum. ∞∑n=1(1en+1n(n+1))

Solution.

Consider the series ∞∑n=1(1e)nand∞∑n=1(1n(n+1)).

Because e>1, the geometric series ∞∑n=1(1e)n converges to 1e−1.

Note that 1n(n+1)=1n−1n+1. Then the partial sum is sn=(1−12)+(12−13)+⋯+(1n−1−1n)+(1n−1n+1)=1−1n+1. Taking the limit, we find that ∞∑n=1(1n(n+1))=limn→∞sn=1.

Therefore, ∞∑n=1(1en+1n(n+1))=ee−1+1=2e−1e−1.

Exercise 10.17 Determine whether the series is convergent or divergent. If it is convergent, find its sum. ∞∑n=1ln(nn+1)

Solution.

Then the partial sum is sn=n∑k=1ln(kk+1)=n∑k=1(lnk−ln(k+1))=ln1−ln(n+1)=−ln(n+1).

Because the limit −limn→∞ln(n+1)=−∞. The series in this question diverges.

Exercise 10.18 Determine whether the series is convergent or divergent. If it is convergent, find its sum. ∞∑n=13n(n+3)

Solution.

The rational expression can be written as a sum of partial fraction: 3n(n+3)=1n−1n+3

Then the partial sum is sn=(1−14)+(12−15)+(13−16)⋯+(1n−2−1n+1)+(1n−1−1n+2)+(1n−1n+3)=1+12+13−1n+1−1n+2−1n+3.

Taking the limit, we find that ∞∑n=13n(n+3)=limn→∞sn=1+12+13=116.

Exercise 10.19 Determine whether the series is convergent or divergent. If it is convergent, find its sum. ∞∑n=1(e1/n−e1/(n+1))

Solution.

The partial sum is sn=(e1−e1/2)+(e1/2−e1/3)⋯+(e1/(n−1)−e1/n)+(e1/n−e1/(n+1))=e−e1/(n+1).

Taking the limit, we find that ∞∑n=13n(n+3)=limn→∞sn=e−e0=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. ∞∑n=1(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∈(−3,−1), the sum of the series is ∞∑n=1(x+2)n=(x+2)⋅11−(x+2)=−x+2x+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. ∞∑n=02nxn

Solution.

The series is the geometric series with r=2x. Therefore, the series converges if |x|>2, or equivalently, x∈(−∞,−2)∪(2,∞). When the series converges, the sum of the series is ∞∑n=0(2x)n=11−2x=xx−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,∞) and let an=f(n). Then the series ∞∑n=1an is convergent if and only if the improper integral ∫∞1f(x)dx is convergent.

As an application, we know exactly when does the p-series ∑1np converge.

Corollary 10.1 The p-series ∞∑n=11np is convergent if p>1 and divergent if p≤1.

When a series ∑an converges, we may use a partial sum sn to estimate the sum s=∑an. The reminder Rn=s−sn can be estimated using integral if the function f with f(n)=an is continuous continuous, positive, decreasing function for x≥n and is convergent: ∫∞n+1f(x)dx≤Rn≤∫∞nf(x)dx

Exercise 10.22 Determine whether the series is convergent or divergent. 1+12√2+13√3+14√4+15√5+⋯

Solution.

Rewrite the radicals using rational exponents, we know that the series is a p-series ∞∑n=11n3/2.

Because p>1, the series converges.

Exercise 10.23 Determine whether the series is convergent or divergent. 15+18+111+114+117+⋯

Solution.

The series can be expressed as 15+18+111+114+117+⋯=∞∑n=112+3n.

Because the improper integral ∫∞113x+2dx=13(limt→∞ln(3x+2)−ln5)=∞, 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. ∞∑n=1n2n3+1

Solution.

Because the improper integral ∫∞1x2x3+1dx=13(limt→∞ln(x3+1)−ln1)=∞, the series diverges by the Integral Test.

Remark: Can you fill in the intermediate steps for the first equality?
Hint: use substitution u=x3+1.

Exercise 10.25 Determine whether the series is convergent or divergent. ∞∑n=11n2+6n+13

Solution.

Because the improper integral ∫∞11x2+6x+13dx=12(limt→∞arctan(t+32)−arctan2)=π−2arctan24, 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 x2+6x+13=(x+3)2+4 and use the substitution x+3=2u.

Exercise 10.26 Determine whether the series is convergent or divergent. ∞∑n=1lnnn3

Solution.

Because the improper integral ∫∞1lnxx3dx=(−limt→∞lnt2t2+ln12)+(−limt→∞14t4+14)=14, 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′(x)=1x3 for the first equality and L’Hospital’s rule for the second equality.

Exercise 10.27 Determine whether the series is convergent or divergent. ∞∑n=1e1/nn2

Solution.

Because the improper integral ∫∞1e1/xx2dx=−limt→∞e1/t+e1=e−1, the series converges by the Integral Test.

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

Exercise 10.28 Find the values p of for which the series is convergent ∞∑n=21n(lnn)p

Solution.

Let u=lnx, then ∫1x(lnx)pdx={ln(lnx)p=11(1−p)(lnx)p−1p≠1

Then

∫∞11x(lnx)pdx{∞p≤10p>1 By the Integral Test, the ∞∑n=21n(lnn)p converges when p>1.

Exercise 10.29 Find the sum of the series ∞∑n=11n3 correct to two decimal places.

Solution.

When using the partial sum sn to approximate the sum, the error (or reminder) R satisfies 12(n+1)2=∫∞n+11x3dx≤R≤∫∞n1x3dx=12n2

Take n=8, we find that 0.006<R<0.008, so the sum s≈s8≈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 ∑an and ∑bn be series with positive terms. Suppose an≥bn for all n. If the series ∑an converges, then the series ∑bn converges. Conversely, if the series ∑bn diverges, then the series ∑an 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 ∑an and ∑bn be series with positive terms. Suppose limn→∞anbn=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 ∑bn converges, then ∑an converges.
3. If c=∞ and ∑bn diverges, then ∑an diverges.

When using the limit comparison test, it’s better to let ∑an to be the given series and take an as the numerator in the limit limn→∞anbn.

Exercise 10.30 Determine whether the series converges or diverges. ∞∑n=2n3n4−1

Solution.

Because the limit limn→∞n3n4−11n=limn→∞n4n4−1=1 and the series ∞∑n=21n diverges, by the limit comparison test, the series ∞∑n=2n3n4−1 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 bn.

Exercise 10.31 Determine whether the series converges or diverges. ∞∑n=1n−1n2√n

Solution.

The question of highest degree terms in the fraction is nn2√n=1n3/2.

Because limn→∞n−1n2√n1n3/2=limn→∞n−1n=1 and the series ∞∑n=21n3/2 converges, by the limit comparison test, the series ∞∑n=2n3n4−1 converges.

Exercise 10.32 Determine whether the series converges or diverges. ∞∑n=19n3+10n

Solution.

We compare the given series with the geometric series ∞∑n=1(910)n.

Because limn→∞9n3+10n9n10n=limn→∞3+10n10n=1 and the series ∞∑n=2(910)n converges, by the limit comparison test, the series ∞∑n=19n3+10n converges.

Exercise 10.33 Determine whether the series converges or diverges. ∞∑k=1ksin2k1+k3

Solution.

Because ∞∑k=11k2 converges and limn→∞k1+k31k2=1 the series ∞∑k=1k1+k3 converges.

Because sin2x≤1, by the comparison test, the series ∞∑k=1ksin2k1+k3 converges.

Exercise 10.34 Determine whether the series converges or diverges. ∞∑n=113√3n4+1

Solution.

Because ∞∑n=113√n4 converges and limn→∞13√3n4+113√n4=13√3 the series ∞∑n=113√3n4+1 converges.

Exercise 10.35 Determine whether the series converges or diverges. ∞∑n=1e1/nn

Solution.

Because ∞∑n=11n diverges and limn→∞=e1/nn1n=1. The series ∞∑n=1e1/nn diverges by the limit comparison test.

Exercise 10.36 Determine whether the series converges or diverges. ∞∑n=11n!

Solution.

Because ∞∑n=11n2 converges. By the limit comparison test, the series ∞∑n=11n(n−1) converges.

Note that 1n!≤1n(n−1) for n≥2. Then the series ∞∑n=11n! converges by the comparison test.

Exercise 10.37 Determine whether the series converges or diverges. ∞∑n=1sin(1n)

Solution.

Because ∞∑n=11n diverges and limn→∞=sin(1n)1n=limx→0+=sinxx=1. The series ∞∑n=1sin(1n) diverges by the limit comparison test.

Exercise 10.38 Determine whether the series converges or diverges. ∞∑n=27lnn

Solution.

Because ∞∑n=11n diverges and limn→∞=7lnn1n=∞. The series ∞∑n=11lnn diverges.

Exercise 10.39 Determine whether the series converges or diverges. ∞∑n=1lnnn2

Solution.

Because ∞∑n=11n3/2 converges and limn→∞=lnnn21n3/2=limn→∞lnnn1/2=limn→∞12n1/2=0. The series ∞∑n=1lnnn2 converges.

Remark: You may find that the limit comparison test is inconclusive if bn=1n or 1n2. But taking bn=1/np 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 ∞∑n=1(−1)n−1bn=b1−b2+b3−b4+b5−b6+⋯bn>0 satisfies the following two conditions: (i) bn+1≤bn for all n and (ii) limn→∞bn=0, then the series converges.

To show that f(n)=bn is decreasing, we often show that the sufficient condition f′(x)<0 holds.

Exercise 10.40 Determine whether the alternating series converges or diverges. 1ln2−1ln3+1ln4−1ln5+1ln6−⋯

Solution.

Because (1lnx)′=−1x(lnx)2<0 for x>2 and limn→∞1lnn=0.

Then the given alternating series converges.

Exercise 10.41 Determine whether the alternating series converges or diverges. ∞∑n=0(−1)2n+1√n+1

Solution.

Because (−1)2n+1=−1 for all n. The series is not an alternating series. Comparing with the p-series ∞∑n=11n1/2, we know that ∞∑n=01√n+1 diverges. So does ∞∑n=0(−1)2n+1√n+1=−∞∑n=01√n+1.

Exercise 10.42 Determine whether the alternating series converges or diverges. ∞∑n=1(−1)n3√n2n+3

Solution.

Because (3√x2x+3)′=13x−2/3(2x+3)−2x1/3(2x+3)2=(3−4x)33√x2(2x+3)2<0 for x>1 and limn→∞3√n2n+3=0.

Then the given alternating series converges.

Exercise 10.43 Determine whether the alternating series converges or diverges. ∞∑n=1(−1)n+1ne−n

Solution.

Because (xe−x)′=e−x(1−x)<0 for x>2 and limn→∞ne−n=limn→∞nen=0.

Then the given alternating series converges.

Exercise 10.44 Determine whether the alternating series converges or diverges. ∞∑n=1(−1)n−12arctann

Solution.

Because limn→∞2arctann=π. The alternating series diverges by the divergence test.

Exercise 10.45 Determine whether the alternating series converges or diverges. ∞∑n=1(−1)nsin(πn)

Solution.

Because (sin(πx))′=−πcos(πx)x2<0 for x>2 and limn→∞sin(πn)=limx→0+sinx=0.

The alternating series converges by the alternating convergence test.

Exercise 10.46 Determine whether the alternating series converges or diverges. ∞∑n=1(−1)nnnn!

Solution.

Because nnn!>1 for all n. Then the divergence test theorem, the series diverges.

Exercise 10.47 Determine whether the alternating series converges or diverges. ∞∑n=1(−1)n(√n+1−√n)

Solution.

Because (√x+1−√x)′=√x−√x+12√x(x+1)<0 for x>0 and limn→∞(√n+1−√n)=limn→∞1√n+1+√n=0.

The alternating series converges by the alternating convergence test.

10.6 Absolute Convergence and Tests

A series ∑an is called absolutely convergent if the series of absolute values ∑|an| is convergent.

A series ∑an is called conditionally convergent if it is convergent but not absolutely convergent.

By comparing ∑(an+|an|) with ∑(2|an|), we draw the following conclusion.

Theorem 10.9 If a series ∑an 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 limn→∞|an+1an|=L<1, then the series ∞∑n=1an is absolutely convergent.
   
3. If limn→∞|an+1an|=L>1, then the series ∞∑n=1an is divergent.

1. Theorem 10.11 (The Root Test)
2. If limn→∞n√|an|=L<1, then the series n∑n=1an is absolutely convergent.
   
3. If limn→∞n√|an|=L>1, then the series n∑n=1an is divergent.

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

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

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

Exercise 10.48 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞∑n=0(−1)n5n+1

Solution.

The series is an alternating series. Because 15n+1 is decreasing and limn→∞15n+1=0. The series is convergent by the alternating convergence test.

Because limn→∞15n+11n=15 and the harmonic series ∞∑n=01n diverges. The absolute series ∞∑n=015n+1 diverges by the limit comparison test.

Exercise 10.49 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞∑n=0(−3)n(2n+1)!

Solution.

Because limn→∞an+1an=limn→∞3n+1(2(n+1)+1)!−3n(2n+1)!=limn→∞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. ∞∑n=1sin4n4n

Solution.

Note that |sin4n|4n≤14n.

Because the geometric series limn→∞14n converges, by the comparison test, the absolute series limn→∞|sin4n|4n converges.

Therefore, the series is absolutely convergent.

Exercise 10.51 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞∑n=13−2sinnn2/3−2

Solution.

Note that 1≤3−2sinn≤5 and 3−2sinnn2/3−2≥1n2/3

Because the series ∞∑n=11n2/3 diverges, by the comparison test, the absolute series ∞∑n=13−2sinnn2/3−2 diverges.

Exercise 10.52 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞∑n=2(−1)nlnn

Solution.

Note that lnn<n for n>1. Then 1lnn>1n.

Because the series ∞∑n=11n diverges, by the comparison test, the absolute series ∞∑n=11lnn diverges.

However, by the alternating convergence test, we know that the series ∞∑n=2(−1)nlnn converges.

Therefore, the series is conditionally convergent.

Exercise 10.53 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞∑n=2(−2nn+1)5n

Solution.

Because the limit ∞∑n=1n√(2nn+1)5n=∞∑n=1(2nn+1)5=25>1. By the root test, the series diverges.

Exercise 10.54 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. 1−1⋅33!+1⋅3⋅55!−1⋅3⋅5⋅77!+⋯+(−1)n−11⋅3⋅5⋅⋯⋅(2n−1)(2n−1)!+⋯

Solution.

Because the limit limn→∞1⋅3⋅5⋅⋯⋅(2n+1)(2n+1)!1⋅3⋅5⋅⋯⋅(2n−1)(2n−1)!=limn→∞12n=0. By the ratio test, the series is absolutely convergent.

Exercise 10.55 Determine whether the series is absolutely convergent, conditionally convergent, or divergent. ∞∑n=12n2n!

Solution.

Because the limit limn→∞2(n+1)2(n+1)!2n2n!=limn→∞22n+1n+1=∞. By the ratio test, the series diverges.

10.7 Power series

A power series is a series of the form ∞∑n=0cnxn=c0+c1x+c2x2+c3x3+⋯, where x is a variable and cn are constant numbers called the coefficients of the series.

More general, the series ∞∑n=0cn(x−a)n=c0+c1(x−a)+c2(x−a)2+⋯ 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 ∞∑n=0cn(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 ∞∑n=0cn(x−a)n at endpoints x=a±R should be inspected carefully.

Exercise 10.56 Find the radius of convergence and interval of convergence of the series. ∞∑n=1(−1)nnxn

Solution.

Because the limit limn→∞|an+1||an|=limn→∞(n+1)|x|n+1n|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 limn→∞n=∞ 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. ∞∑n=1(−1)nxnn2

Solution.

Because the limit limn→∞|an+1||an|=limn→∞n2|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 ∞∑n=1(−1)nxnn2 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. ∞∑n=1(−1)nn2xn2n

Solution.

Because the limit limn→∞|an+1||an|=limn→∞(n+1)22n|x|n+1n22n+1|x|n=|x|2.

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

When |x|=2, limn→∞(−1)nn2xn2n 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. ∞∑n=0(−1)nx2n+1(2n+1)!

Solution.

Because the limit limn→∞|an+1||an|=limn→∞|x|2n+3(2n+3)!|x|2n+1(2n+1)!=limn→∞|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=∞ and the interval of convergence is (−∞,∞).

Exercise 10.60 Find the radius of convergence and interval of convergence of the series. ∞∑n=13n(x+4)n√n

Solution.

Because the limit limn→∞|an+1||an|=limn→∞3n+1|x+4|n+1√n+13n|x+4|n√n=limn→∞3|x+4|√n√n+1=3|x+4|.

By ratio test, we know the the series converges if |x+4|<13. So the radius of convergence is R=13.

When x+4=13, the power series is a divergent p-series with p=12.

When x+4=−13, the power series is a convergent alternating series.

Therefore, the interval of convergence is [−13−4,13−4)=[−133,−113).

Exercise 10.61 Find the radius of convergence and interval of convergence of the series. ∞∑n=2bnlnn(x−a)n,b>0

Solution.

Because the limit limn→∞|an+1||an|=limn→∞bn+1|x−a|n+1ln(n+1)bn|x−a|nlnn=limn→∞b|x−a|lnnln(n+1)=b|x−a|.

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

When x−a=1b, the power series is a divergent by comparing with the p-series ∑n→∞1n.

When x−a=−1b, the power series is a convergent alternating series.

Therefore, the interval of convergence is [−1b+a,1b+a)=[ab−1b,−ab+1b).

Exercise 10.62 Find the radius of convergence and interval of convergence of the series. ∞∑n=1n!(2x−1)n

Solution.

Because the limit limn→∞|an+1||an|=limn→∞(n+1)!|2x−1|n+1n!|2x−1|n={0if  x=12∞otherwise.

By ratio test, we know the the series converges only if x=12. So the radius of convergence is R=0.

Therefore, the interval of convergence is a one-point set {12}.

Exercise 10.63 Find the radius of convergence and interval of convergence of the series. ∞∑n=2x2nn(lnn)2

Solution.

Because the limit limn→∞|an+1||an|=limn→∞|x|2n+2(n+1)(ln(n+1))2|x|2nn(lnn)2=limn→∞n(lnn)2|x|2(n+1)(ln(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 ∞∑n=0cn(x−a)n, we may define a function f(x)=∞∑n=0cn(x−a)n over the interval of convergence.

For example, 11−x=1+x+x2+x3+⋯=∞∑n=0xnfor  |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\limits_{n=0}^{\infty} c_{n}(x-a)^{n} has radius of convergence R, then the function f defined by f(x)=\sum\limits_{n=0}^{\infty} c_{n}(x-a)^{n} is differentiable over the interval (a-R, a+R). Moreover, \frac{\mathrm{d}}{\mathrm{d} x} f(x)= \frac{\mathrm{d}}{\mathrm{d} x}\left[\sum\limits_{n=0}^{\infty} c_{n}(x-a)^{n}\right]=\sum\limits_{n=0}^{\infty} \frac{\mathrm{d}}{\mathrm{d} x}\left[c_{n}(x-a)^{n}\right] and \int f(x)\mathrm{d} x=\int\left[\sum\limits_{n=0}^{\infty} c_{n}(x-a)^{n}\right]\mathrm{d} x=\sum\limits_{n=0}^{\infty} \int\left[c_{n}(x-a)^{n}\mathrm{d} x\right]

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

Solution.

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

We obtain \begin{aligned} \frac{1}{2-x}=&\frac12\cdot\frac{1}{1-\frac x2}\\ =&\frac12\cdot\left(1+\frac x2+\left(\frac x2\right)^2+\cdots\right)\\ =&\frac12+\frac x{2^2}+\frac{x^2}{2^3}+\cdots\\ =&\sum_{n=0}^\infty\frac{x^{n}}{2^{n+1}} \end{aligned}

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(x)=\frac{x}{4+x^{2}}

Solution.

A power series expansion can be obtained as follows \begin{aligned} \frac{x}{4+x^{2}}=&\frac x4\cdot\frac{1}{1-\left(-\frac{x^2}{4}\right)}\\ =&\frac x4\cdot\left(1+\left(-\frac{x^2}{4}\right)+\left(-\frac{x^2}{4}\right)^2+\cdots\right)\\ =&\frac x4-\frac{x^3}{4^2}+\frac{x^5}{4^3}-\cdots\\ =&\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{4^{n+1}}. \end{aligned}

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(x)=\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 (\ln(5-x))^\prime=-\frac{1}{5-x}. By taking integral, we have \begin{aligned} \ln(5-x)=&\int -\frac{1}{5-x} \mathrm{d} x\\ =& \int\left(\sum_{n=0}^\infty\frac{x^{n}}{5^{n+1}}\right)\mathrm{d} x\\ =& \sum_{n=0}^\infty\int \frac{x^{n}}{5^{n+1}}\mathrm{d} x\\ =& -\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{aligned} 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(x)=x^{2} \arctan\left(x^{3}\right)

Solution.

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

Note that \begin{aligned} (\arctan(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{aligned}

Then \begin{aligned} \arctan(x^3)=&\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} \qquad \text{because}\quad \arctan 0=0 \end{aligned}

Therefore, x^2\arctan(x^3)\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(x)=\left(\frac{x}{2-x}\right)^{3}

Solution.

A power series representation can be obtained as follows \begin{aligned} \left(\frac{x}{2-x}\right)^{3}=&x^3\left(\frac{1}{2-x}\right)^{3}\\ =&x^3\cdot \frac12\cdot \left(\frac{1}{2-x}\right)^{\prime\prime}\\ =&\frac12 x^3\left(\sum_{n=0}^\infty\frac{x^{n}}{2^{n+1}}\right)^{\prime\prime}\\ =&\frac12 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{aligned}

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}} d t

Solution.

The rational function f(t)=\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{aligned} \int \frac{t}{1-t^8}\mathrm{d} t =&\int\left(\sum_{n=0}^\infty t^{8n+1}\right) \mathrm{d} t\\ =&\sum_{n=0}^\infty \left(\int t^{8n+1}\mathrm{d} t\right) \\ =&\sum_{n=0}^\infty \frac{t^{8n+2}}{8n+2}+C \\ \end{aligned}

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} d x

Solution.

The function f(x)=\arctan x has a power series representation as follows \begin{aligned} \arctan x=&\int\frac{1}{1+x^2}\mathrm{d} x + C_0\\ =& \int \sum_{n=0}^\infty (-1)^nx^{2n} \mathrm{d} x + C_0\\ =& \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1} + C\\ =& \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1}\\ \end{aligned}

Therefore, the integral has a power series representation \begin{aligned} \int \frac{\arctan x}{x} d x =&\int\left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{2n+1}\right) \mathrm{d} t\\ =&\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)^2}+C \\ \end{aligned}

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}} d x

Solution.

The indefinite integral has a power series representation as follows \begin{aligned} \int\frac1{1+x^5}\mathrm{d} x =& \int \sum_{n=0}^\infty (-1)^nx^{5n} \mathrm{d} x\\ =& \sum_{n=0}^\infty \frac{(-1)^nx^{5n+1}}{5n+1} + C \end{aligned}

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}\frac1{1+x^5}\mathrm{d} x\approx \frac{0.2}{1}-\frac{0.2^6}{6}\approx 0.2- 0.000011=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(x)=\sum_{n=0}^\infty c_n(x-a)^n\qquad |x-a|<R, then the coefficients c_n are determined by the formula c_n=\frac{f^{(n)}(a)}{n!}. In other words, if f has power series expansion, then it must be f(x)=f(a)+\frac{f^{\prime} (a)}{1!}(x-a) + \frac{f^{\prime\prime} (a)}{2!}(x-a)^2 + \frac{f^{\prime\prime\prime} (a)}{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^{(k)}(a)}{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\neq a. Even it converges for x\neq a, the sum of the Taylor series may not be f(x).

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(x) for x\neq a can be given using the reminder of the Taylor series.

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

If f is the sum of its Taylor series, then \lim\limits_{n\to\infty} T_n(x)=f(x) or equivalently \lim\limits_{n\to\infty}(f(x)-T_n(x))=0. We call the difference R_n(x)=f(x)-T_n(x) 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(x)= \sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k + R_n(x), where R_n(x)=\frac{1}{n!}\int_a^x(x-t)^nf^{(n+1)}(t)\mathrm{d} t = \frac{f^{(n+1)}(c)}{(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(x) can be proved by induction. The second equality for R_n(x) is obtained by the weighted mean-value theorem for integral. The reminder R_n(x)=\frac{f^{(n+1)}(c)}{(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(x).

Theorem 10.16 If |f^{(n+1)}(x)|\leq M for |x-a|\leq R and \lim_{n\to \infty} \frac{M}{(n+1)!}|x-a|^{n+1}=0 \qquad \text{for}\quad|x-a|\leq R. Then f(x)=\sum_{n=0}^\infty c_n(x-a)^n\qquad \text{for}\quad |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
\frac1{1-x}=\sum\limits_{n=0}^\infty x^n=1+x+x^2+x^3+\cdots	R = 1	(-1, 1)
e^x=\sum\limits_{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\limits_{n=0}^\infty\frac{(-1)^nx^{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\limits_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots	R = \infty	(-\infty, \infty)
\arctan x=\sum\limits_{n=0}^\infty\frac{(-1)^nx^{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\limits_{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\limits_{n=0}^\infty{k\choose 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(x)=\ln x, \qquad a=1.

Solution.

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

Therefore, the Taylor series of the function f(x)=\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\limits_{n=1}^\infty\frac{(-1)^{n+1}(x-1)}{n} using the power series representation \ln(1+x)=\sum\limits_{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(x)=\arctan(x^2)

Solution.

Using the power series expansion \arctan x=\sum_{n=0}^\infty(-1)^{n}\frac{x^{2n+1}}{2n+1} we get \arctan(x^2)=\sum_{n=0}^\infty(-1)^{n}\frac{x^{4n+2}}{2n+1}. By Theorem 10.14, we know that \sum\limits_{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(x)=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(x)=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(x)=\sin(\pi x)

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(\pi x)=\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(x)=x^2e^{2x}

Solution.

Using the power series expansion e^x=\sum_{n=0}^\infty\frac{x^n}{n!} we get the Maclaurin series x^2e^{2x}=x^2\sum_{n=0}^\infty\frac{2^nx^{n}}{n!}=\sum_{n=0}^\infty\frac{2^nx^{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(x)=\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{aligned} \frac{x}{\sqrt{4+x}}=&\frac{x}{2}\cdot(1+\frac{x}{2})^{-1/2}\\ =& \frac{x}{2}\left(1-\frac{1}{2} \left(\frac x2\right)+\frac{1\cdot 3}{2^2\cdot 2!}\left(\frac x2\right)^2-\frac{1\cdot 3 \cdot 5}{2^3\cdot 3!}\left(\frac x2\right)^3+\cdots\right)\\ =& \frac{x}{2}\left(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\right)\\ =& \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{aligned}

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

Solution.

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

Taking integral of both sides, we get \begin{aligned} \int\frac{\cos x-1}x\mathrm{d} x =&\int \left(\sum\limits_{n=1}^\infty\frac{(-1)^nx^{2n-1}}{(2n)!}\right)\mathrm{d} x\\ =& \sum\limits_{n=1}^\infty\left(\int \frac{(-1)^nx^{2n-1}}{(2n)!} \mathrm{d} x \right)\\ =&\sum\limits_{n=1}^\infty\frac{(-1)^nx^{2n}}{2n(2n)!}. \end{aligned}

Exercise 10.79 Evaluate the limit \lim\limits_{x\to 0}\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}=-\frac13+\frac{x^2}{5}-\frac{x^4}{7}+\cdots Using the fact that -\frac13+\frac{x^2}{5}-\frac{x^4}{7}+\cdots is a continuous function, we find that \begin{aligned} \lim\limits_{x\to 0}\frac{x-\arctan x }{x^3} =&\lim_{x\to 0}\left(-\frac13+\frac{x^2}{5}-\frac{x^4}{7}+\cdots\right)\\ =&-\frac13+\frac{0^2}{5}-\frac{0^4}{7}+\cdots\\ =&-\frac13. \end{aligned}

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 \frac1{2} + \frac1{2\cdot 2^2} + \frac1{3\cdot 2^3} + \frac1{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=\frac12 into both sides, we get \frac1{2} + \frac1{2\cdot 2^2} + \frac1{3\cdot 2^3} + \frac1{4\cdot 2^4} + \cdots =-\ln(1-\frac12)=\ln 2.