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Infinite Series

Module by: Lawrence Baggett. E-mail the author

Summary: Probably the most interesting and important examples of sequences are those that arise as the partial sums of an infinite series. In fact, it will be infinite series that allow us to explain such things as trigonometric and exponential functions.

Probably the most interesting and important examples of sequences are those that arise as the partial sums of an infinite series. In fact, it will be infinite series that allow us to explain such things as trigonometric and exponential functions.

Definition 1:

Let {an}0{an}0 be a sequence of real or complex numbers. By the infinite seriesanan we mean the sequence {SN}{SN} defined by

S N = n = 0 N a n . S N = n = 0 N a n .
(1)

The sequence {SN}{SN} is called the sequence of partial sums of the infinite series an,an, and the infinite series is said to be summable to a number S,S, or to be convergent, if the sequence {SN}{SN} of partial sums converges to S.S.The sum of an infinite series is the limit of its partial sums.

An infinite series anan is called absolutely summable or absolutely convergent if the infinite series |an||an| is convergent.

If anan is not convergent, it is called divergent. If it is convergent but not absolutely convergent, it is called conditionally convergent.

A few simple formulas relating the anan's and the SNSN's are useful:

S N = a 0 + a 1 + a 2 + ... + a N , S N = a 0 + a 1 + a 2 + ... + a N ,
(2)
S N + 1 = S N + a N + 1 , S N + 1 = S N + a N + 1 ,
(3)

and

S M - S K = n = K + 1 M a n = a K + 1 + a K + 2 + ... + A M , S M - S K = n = K + 1 M a n = a K + 1 + a K + 2 + ... + A M ,
(4)

for M>K.M>K.

REMARK Determining whether or not a given infinite series converges is one of the most important and subtle parts of analysis. Even the first few elementary theorems depend in deep ways on our previous development, particularly the Cauchy criterion.

Theorem 1

Let {an}{an} be a sequence of nonnegative real numbers. Then the infinite series anan is summable if and only if the sequence {SN}{SN} of partial sums is bounded.

Proof

If anan is summable, then {SN}{SN} is convergent, whence bounded according to (Reference). Conversely, we see from the hypothesis that each an0an0 that {SN}{SN} is nondecreasing (SN+1=SN+aN+1SNSN+1=SN+aN+1SN). So, if {SN}{SN} is bounded, then it automatically converges by (Reference), and hence the infinite series anan is summable.

The next theorem is the first one most calculus students learn about infinite series. Unfortunately, it is often misinterpreted, so be careful! Both of the proofs to the next two theorems use (Reference), which again is a serious and fundamental result about the real numbers. Therefore, these two theorems must be deep results themselves.

Theorem 2

Let anan be a convergent infinite series. Then the sequence {an}{an} is convergent, and liman=0.liman=0.

Proof

Because anan is summable, the sequence {SN}{SN} is convergent and so is a Cauchy sequence. Therefore, given an ϵ>0,ϵ>0, there exists an N0N0 so that |Sn-Sm|<ϵ|Sn-Sm|<ϵ whenever both nn and mN0.mN0. If n>N0,n>N0, let m=n-1.m=n-1. We have then that |an|=|Sn-Sm|<ϵ,|an|=|Sn-Sm|<ϵ, which completes the proof.

REMARK Note that this theorem is not an “if and only if” theorem. The harmonic series (part (b) of Exercise 2 below) is the standard counterexample. The theorem above is mainly used to show that an infinite series is not summable. If we can prove that the sequence {an}{an} does not converge to 0, then the infinite series anan does not converge. The misinterpretation of this result referred to above is exactly in trying to apply the (false) converse of this theorem.

Theorem 3

If anan is an absolutely convergent infinite series of complex numbers, then it is a convergent infinite series. (Absolute convergence implies convergence.)

Proof

If {SN}{SN} denotes the sequence of partial sums for an,an, and if {TN}{TN} denotes the sequence of partial sums for |an|,|an|, then

| S M - S N | = | n = N + 1 M a n | n = N + 1 M | a n | = | T M - T N | | S M - S N | = | n = N + 1 M a n | n = N + 1 M | a n | = | T M - T N |
(5)

for all NN and M.M. We are given that {TN}{TN} is convergent and hence it is a Cauchy sequence. So, by the inequality above, {SN}{SN} must also be a Cauchy sequence. (If |TN-TM|<ϵ,|TN-TM|<ϵ, then |SN-SM|<ϵ|SN-SM|<ϵ as well.) This implies that anan is convergent.

Exercise 1: The Infinite Geometric Series

Let zz be a complex number, and define a sequence {an}{an} by an=zn.an=zn. Consider the infinite series an.an. Show that n=0ann=0an converges to a number SS if and only if |z|<1.|z|<1. Show in fact that S=1/(1-z),S=1/(1-z), when |z|<1.|z|<1.

HINT: Evaluate explicitly the partial sums SN,SN, and then take their limit. Show that SN=1-zN+11-z.SN=1-zN+11-z.

Exercise 2

  1. Show that n=11n(n+1)n=11n(n+1) converges to 1,1, by computing explicit formulas for the partial sums. HINT: Use a partial fraction decomposition for the anan's.
  2. (The Harmonic Series.) Show that n=11/nn=11/n diverges by verifying that S2k>k/2.S2k>k/2. HINT: Group the terms in the sum as follows,
    1+12+(13+14)+(15+16+17+18)+(19+110+...+116)+...,1+12+(13+14)+(15+16+17+18)+(19+110+...+116)+...,
    (6)
    and then estimate the sum of each group. Remember this example as an infinite series that diverges, despite the fact that is terms tend to 0.

The next theorem is the most important one we have concerning infinite series of numbers.

Theorem 4: Comparison Test

Suppose {an}{an} and {bn}{bn} are two sequences of nonnegative real numbers for which there exists a positive integer MM and a constant CC such that bnCanbnCan for all nM.nM. If the infinite series anan converges, so must the infinite series bn.bn.

Proof

We will show that the sequence {TN}{TN} of partial sums of the infinite series bnbn is a bounded sequence. Then, by Theorem 1, the infinite series bnbn must be summable.

Write SNSN for the NNth partial sum of the convergent infinite series an.an. Because this series is summable, its sequence of partial sums is a bounded sequence. Let B be a number such that SNBSNB for all N.N. We have for all N>MN>M that

r c l T N = n = 1 N b n = n = 1 M b n + n = M + 1 N b n n = 1 M b n + n = M + 1 N C a n = n = 1 M b n + C n = M + 1 N a n n = 1 M b n + C n = 1 N a n n = 1 M b n + C S N n = 1 M b n + C B , r c l T N = n = 1 N b n = n = 1 M b n + n = M + 1 N b n n = 1 M b n + n = M + 1 N C a n = n = 1 M b n + C n = M + 1 N a n n = 1 M b n + C n = 1 N a n n = 1 M b n + C S N n = 1 M b n + C B ,
(7)

which completes the proof, since this final quantity is a fixed constant.

Exercise 3

  1. Let {an}{an} and {bn}{bn} be as in the preceding theorem. Show that if bnbn diverges, then anan also must diverge.
  2. Show by example that the hypothesis that the anan's and bnbn's of the Comparison Test are nonnegative can not be dropped.

Exercise 4: The Ratio Test

Let {an}{an} be a sequence of positive numbers.

  1. If lim supan+1/an<1,lim supan+1/an<1, show that anan converges. HINT: If lim supan+1/an=α<1,lim supan+1/an=α<1, let ββ be a number for which α<β<1.α<β<1. Using part (a) of (Reference), show that there exists an NN such that for all n>Nn>N we must have an+1/an<β,an+1/an<β, or equivalently an+1<βan,an+1<βan, and therefore aN+k<βkaN.aN+k<βkaN. Now use the comparison test with the geometric series βk.βk.
  2. If lim infan+1/an>1,lim infan+1/an>1, show that anan diverges.
  3. As special cases of parts (a) and (b), show that {an}{an} converges if limnan+1/an<1,limnan+1/an<1, and diverges if limnan+1/an>1.limnan+1/an>1.
  4. Find two examples of infinite series' anan of positive numbers, such that liman+1/an=1liman+1/an=1 for both examples, and such that one infinite series converges and the other diverges.

Exercise 5

  1. Derive the Root Test: If {an}{an} is a sequence of positive numbers for which lim supan1/n<1,lim supan1/n<1, then anan converges. And, if lim infan1/n>1,lim infan1/n>1, then anan diverges.
  2. Let rr be a positive integer. Show that 1/nr1/nr converges if and only if r2.r2. HINT: Use Exercise 2 and the Comparison Test for r=2.r=2.
  3. Show that the following infinite series are summable.
    1/(n2+1),n/2n,an/n!,1/(n2+1),n/2n,an/n!,
    (8)
    for aa any complex number.

Exercise 6

Let {an}{an} and {bn}{bn} be sequences of complex numbers, and let {SN}{SN} denote the sequence of partial sums of the infinite series an.an. Derive the Abel Summation Formula:

n = 1 N a n b n = S N b N + n = 1 N - 1 S n ( b n - b n + 1 ) . n = 1 N a n b n = S N b N + n = 1 N - 1 S n ( b n - b n + 1 ) .
(9)

The Comparison Test is the most powerful theorem we have about infinite series of positive terms. Of course, most series do not consist entirely of positive terms, so that the Comparison Test is not enough. The next theorem is therefore of much importance.

Theorem 5: Alternating Series Test

Suppose {a1,a2,a3,...}{a1,a2,a3,...} is an alternating sequence of real numbers; i.e., their signs alternate. Assume further that the sequence {|an|}{|an|} is nonincreasing with 0=lim|an|.0=lim|an|. Then the infinite series anan converges.

Proof

Assume, without loss of generality, that the odd terms a2n+1a2n+1 of the sequence {an}{an} are positive and the even terms a2na2n are negative. We collect some facts about the partial sums SN=a1+a2+...+aNSN=a1+a2+...+aN of the infinite series an.an.

  1. Every even partial sum S2NS2N is less than the following odd partial sum S2N+1=S2N+a2N+1,S2N+1=S2N+a2N+1, And every odd partial sum S2N+1S2N+1 is greater than the following even partial sum S2N+2=S2N+1+a2N+2.S2N+2=S2N+1+a2N+2.
  2. Every even partial sum S2NS2N is less than or equal to the next even partial sum S2N+2=S2N+a2N+1+a2N+2,S2N+2=S2N+a2N+1+a2N+2, implying that the sequence of even partial sums {S2N}{S2N} is nondecreasing.
  3. Every odd partial sum S2N+1S2N+1 is greater than or equal to the next odd partial sum S2N+3=S2N+1+a2N+2+a2N+3,S2N+3=S2N+1+a2N+2+a2N+3, implying that the sequence of odd partial sums {S2N+1}{S2N+1} is nonincreasing.
  4. Every odd partial sum S2N+1S2N+1 is bounded below by S2.S2. For, S2N+1>S2NS2.S2N+1>S2NS2. And, every even partial sum S2NS2N is bounded above by S1.S1. For, S2N<S2N+1S1.S2N<S2N+1S1.
  5. Therefore, the sequence {S2N}{S2N} of even partial sums is nondecreasing and bounded above. That sequence must then have a limit, which we denote by Se.Se. Similarly, the sequence {S2N+1}{S2N+1} of odd partial sums is nonincreasing and bounded below. This sequence of partial sums also must have a limit, which we denote by So.So.

Now

S o - S e = lim S 2 N + 1 - lim S 2 N = lim ( S 2 N + 1 - S 2 N ) = lim a 2 N + 1 = 0 , S o - S e = lim S 2 N + 1 - lim S 2 N = lim ( S 2 N + 1 - S 2 N ) = lim a 2 N + 1 = 0 ,
(10)

showing that Se=So,Se=So, and we denote this common limit by S.S. Finally, given an ϵ>0,ϵ>0, there exists an N1N1 so that |S2N-S|<ϵ|S2N-S|<ϵ if 2NN1,2NN1, and there exists an N2N2 so that |S2N+1-S|<ϵ|S2N+1-S|<ϵ if 2N+1N2.2N+1N2. Therefore, if Nmax(N1,N2),Nmax(N1,N2), then |SN-S|<ϵ,|SN-S|<ϵ, and this proves that the infinite series converges.

Exercise 7: The Alternating Harmonic Series

  1. Show that n=1(-1)n/nn=1(-1)n/n converges, but that it is not absolutely convergent.
  2. Let {an}{an} be an alternating series, as in the preceding theorem. Show that the sum S=anS=an is trapped between SNSN and SN+1,SN+1, and that |S-SN||aN|.|S-SN||aN|.
  3. State and prove a theorem about “eventually alternating infinite series.”
  4. Show that zn/nzn/n converges if and only if |z|1,|z|1, and z1.z1. HINT: Use the Abel Summation Formula to evaluate the partial sums.

Exercise 8

Let s=p/qs=p/q be a positive rational number.

  1. For each x>0,x>0, show that there exists a unique y>0y>0 such that ys=x;ys=x; i.e., yp=xq.yp=xq.
  2. Prove that 1/ns1/ns converges if s>1s>1 and diverges if s1.s1. HINT: Group the terms as in part (b) of Exercise 2.

Theorem 6: Test for Irrationality

Let xx be a real number, and suppose that {pN/qN}{pN/qN} is a sequence of rational numbers for which x=limpN/qNx=limpN/qN and xpN/qNxpN/qN for any N.N. If limqN|x-pN/qN|=0,limqN|x-pN/qN|=0, then xx is irrational.

Proof

We prove the contrapositive statement; i.e., if x=p/qx=p/q is a rational number, then limqN|x-pN/qN|0.limqN|x-pN/qN|0. We have

x - p N / q N = p / q - p N / q N = p q N - q p N q q N . x - p N / q N = p / q - p N / q N = p q N - q p N q q N .
(11)

Now the numerator pqN-qpNpqN-qpN is not 0 for any N.N. For, if it were, then x=p/q=pN/qN,x=p/q=pN/qN, which we have assumed not to be the case. Therefore, since pqN-qpNpqN-qpN is an integer, we have that

| x - p N / q N | = | p q N - q p N q q N | 1 | q q N | . | x - p N / q N | = | p q N - q p N q q N | 1 | q q N | .
(12)

So,

q N | x - p N / q N | 1 | q | , q N | x - p N / q N | 1 | q | ,
(13)

and this clearly does not converge to 0.

Exercise 9

  1. Let x=n=0(-1)n/2n.x=n=0(-1)n/2n. Prove that xx is a rational number.
  2. Let y=n=0(-1)n/2n2.y=n=0(-1)n/2n2. Prove that yy is an irrational number. HINT: The partial sums of this series are rational numbers. Now use the preceding theorem and part (b) of Exercise 7.

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