1 Using Convergence Tests 2 the Cauchy Condensation Test

1 Using convergence tests

1. Decide whether each of the following series converge or diverge: ∞ ∞ ∞ X sin n X (−1)n X n2 ; √ ; . n2 4 + n (log log n)log n n=1 n=1 n=3 [Hint for the last one: Rewrite the denominator using ab = eb log a.] 2. Find all real numbers x for which the following series converge: ∞ ∞ X n2xn X xn(1 − x) ; 5n n n=0 n=1

2 The Cauchy condensation test

In this section, we state and prove the Cauchy condensation test, an alternative to the integral test which is more “elementary” in the sense that we do not need to develop any further theory than that of sequences and series. (In particular, we do not need to develop integration.)

Let (an) be a sequence which is decreasing and converges to 0. (For such a sequence, we may write an & 0.) The Cauchy condensation test states that ∞ ∞ X X k an converges ⇐⇒ 2 a2k converges. n=1 k=0 The next two problems give a proof of this result. 1. Let N K X X k SN = an (N ∈ N) and TK = 2 a2k (K ∈ N ∪ {0}). n=1 k=0 Show for all K ≥ 0 that 1 S K+1 ≤ T and (T − a ) ≤ S K − a . 2 −1 K 2 K 1 2 1 2. Use the previous bounds together with the convergence of bounded monotone sequences to prove the validity of the Cauchy condensation test. The remaining problems apply the Cauchy condensation test to various series.

P p 3.( p-test) Fix p ∈ R. Show that 1/n converges if and only if p > 1. 4. The p-test suggests that the harmonic series P 1/n is more-or-less the smallest diverging series, as any power of n in the denominator greater than 1 gives a converging series. ∞ X 1 However, there are smaller diverging series: show that diverges. n log n n=2

pn 5. For each n ∈ N, let pn denote the n-th prime number. Using the fact that lim = 1, n→∞ n log n P a consequence of the Prime Number Theorem, show that 1/pn diverges.

1 3 The limit comparison test

Throughout, let (an) and (bn) be positive sequences. P 1. Show that if bn converges and a  lim sup n < +∞, bn P then an converges. P 2. Show that if bn diverges and a  lim inf n > 0, bn P then an diverges. 3. Show that if a  0 < lim n < +∞, n→∞ bn P P then an converges if and only if bn converges.

4 Summation by parts and alternating series

N X 1. (Summation by parts) Let (an) and (bn) be two sequences, then let AN = an be the n=1 P N-th partial sum of an. Use induction on N to show that for all M,N ∈ N with M ≤ N, N N−1 X X anbn = (AN bN − AM−1bM ) + An(bn − bn+1). n=M n=M

[When N = M, the sum on the right hand side equals 0. Also, we take A0 = 0.] N X 2. (Dirichlet’s test) Let (an) and (bn) be two sequences, then let BN = bn be the N-th n=1 P partial sum of bn. Suppose an & 0, i.e. (an) is decreasing and an → 0, and that the ∞ sequence (BN )N=1 is bounded (but not necessarily convergent). (a) Use summation by parts to show that

N N−1 X X anbn = aN BN + Bn(an − an+1). n=1 n=1 P (b) Show that |Bn(an − an+1)| is absolutely convergent. P (c) Conclude that anbn converges. 3. Use Dirichlet’s test to give another proof of the alternating series test. P 4. (Alternating series error bound) Let an be a converging alternating series and let AN P be its N-th partial sum. Show that |AN − an| < |aN+1|.

2 5 Series convergence and rate of decay

In this section, we investigate the relationship between the speed at which a (positive) sequence P (an) converges to 0 and the convergence of the series an.

Let (an) and (bn) be two positive sequences converging to 0. We say that (an) decays faster than (bn) if an/bn → 0. In this case, we can also say that (bn) decays slower than (an). 1. Prove the following claims: (a) If p, q > 0, then 1/np decays faster than 1/nq if and only if p > q. (b) If p > 0 and 0 < r < 1, then rn decays faster than 1/np. (c) If 0 < r, s < 1, then rn decays faster than sn if and only if r < s. (d) If 0 < r < 1, then 1/n! decays faster than rn. 2. In this problem, we show that there is no “fastest-decaying diverging series,” i.e. given any P positive sequence (bn) converging to 0 for which bn diverges, there is a positive sequence P (an) converging to 0 which decays faster than (bn) yet still has an diverging. P (a) Let (bn) be a positive sequence converging to 0 for which bn diverges, and define

bn an = Pn . k=1 bk

Show that (an) is a positive sequence converging to 0 which decays faster than (bn). P (b) Let BN be the N-th partial sum of bn. Explain why we can choose natural numbers

N1 < N2 < N3 < ··· so that BNk+1 > 2BNk for all k ∈ N. (c) Show that Nk+1 X BNk+1 − BNk 1 an ≥ > BNk+1 2 n=Nk+1 P for all k ∈ N. Conclude that an diverges. 3. In this problem, we show that there is no “slowest-decaying convergent series.” P P (a) Let an be a positive sequence converging to 0 for which an converges, and define a b = n . n pP∞ k=n ak

Show that (bn) is a positive sequence converging to 0 and (an) decays faster than (bn). P∞ P (b) Let TN = n=N an be the N-th tail of an. Explain why the sequence defined recursively by N1 = min{N ∈ N | TN ≤ 1/2} and

Nk+1 = min{N ∈ N | N > Nk and TN ≤ TNk /2}

is well-defined and strictly increasing. Note that TNk /2 ≤ TNk+1−1 ≤ TNk . (c) Show that

Nk+1−1 r X TNk − TNk+1−1 TN √ b ≤ + b ≤ k + a ≤ 2pT ≤ 21−k/2 n p Nk+1−1 Nk+1−1 Nk TNk+1−1 2 n=Nk P for all k ∈ N. Conclude that bn converges.

3 6 Rearrangement of series

In this section, we investigate the effect of rearranging terms in a series.

1. Let SN and TN be the N-th partial sums of the two series 1 1 1 1 1 1 1 1 1 1 − + − + − · · · and 1 + − + + − + ··· , 2 3 4 5 3 2 5 7 4 which have the same terms but in different orders. Let 1 1 1 H = 1 + + + ··· + N 2 3 N be the N-th partial sum of the harmonic series.

(a) Show that (SN ) converges to a limit S. (b) Show that 1 1 S = H − H and T = H − H − H . 2N 2N N 3N 4N 2 2N 2 N

(c) Deduce that TN → 3S/2. P 2. Let an be an absolutely convergent series and let σ : N → N be a permutation of N, i.e. a bijective function N → N. (Informally, σ is a rearrangement of the natural numbers.) In P P this problem, we show that aσ(n) = an, i.e. rearranging the terms of an absolutely convergent series has no effect on its convergence and value. Fix  > 0.

∞ X (a) Explain why there exists N1 be such that |an| < .

n=N1+1

(b) Explain why there exists N2 such that {1,...,N1} ⊆ {σ(1), . . . , σ(N)} for N ≥ N2. P P (c) Let AN and BN be the N-th partial sums of an and aσ(n). Show that

|AN1 − A| <  and |BN − AN1 | < 

for all N ≥ N2.

(d) Deduce that BN → A.

Thus we see that absolute convergence gives us more than just a convenient way to test ordinary convergence: It assures us that we can rearrange terms however we want without changing the value of a series, which can be useful for evaluating them. For much of the rest of the section, we consider series which converge but do not absolutely converge. Such series are called conditionally convergent, and what we will ultimately show is that rearranging the terms of a conditionally convergent series can cause it to fail to converge, and even if the rearranged series remains convergent, it can take on any real value.

P 3. Adapt the argument in Problem 2 to show that if an is a series with non-negative terms P P and an = +∞, then aσ(n) = +∞ for any permutation σ : N → N. Moreover, this conclusion holds even if there are negative terms as long as the sum of all negative terms is finite. (In particular, the conclusion holds if there are only finitely many negative terms.)

4 P 4. Let an be a convergent series, and let (bk) = (ank ) and (cl) = (anl ) be the “subsequences” of non-negative and negative terms, respectively. Prove that exactly one of the following two cases must hold: P P P (I) k bk and l cl are both finite. In this case, an is absolutely convergent with

∞ ∞ X X X X X X an = bk + cl and |an| = bk − cl. n=1 k l n=1 k l

(The quotes on “subsequences” are because it is possible for there to only be finitely many non-negative or finitely many negative terms.) P P P (II) k bk and l cl are both infinite. In this case, ak is conditionally convergent. Thus conditionally convergent series have infinitely many positive and negative terms. P 5. Let an be a conditionally convergent series and let (bk) and (cl) be as in Problem 4. If P x ≥ 0 is given, we define a new series by listing the terms of an in the following order:

(1) list terms of (bk), in order, until their sum is greater than x;

(2) list terms of (cl), in order, until the sum of all terms listed so far is smaller than x;

(3) list terms of (bk), in order starting from the first one that was not used earlier, until the sum of all terms listed so far is greater than x;

(4) list terms of (cl), in order starting from the first one that was not used earlier, until the sum of all terms listed so far is less than x; (5) repeat steps (3) and (4) ad infinitum. A similar procedure applies to x < 0. P Show that this new series converges to x. Hence if an is a conditionally convergent series P and x ∈ R is arbitrary, we can find a permutation σ : N → N for which aσ(n) = x. P 6. Adapt the construction in Problem 5 to prove that if an is a conditionally convergent series, then P (a) there is a permutation σ : N → N such that the N-th partial sums of aσ(n) do not have a limit; P (b) there is a permutation σ : N → N such that aσ(n) = +∞; P (c) there is a permutation σ : N → N such that aσ(n) = −∞.

5 7 The monotone convergence theorem for series

Let (ak,n) be a doubly-indexed sequence and suppose for each fixed n that an = lim ak,n is a k→∞ finite real number. The monotone convergence theorem is a result on when we can say

∞ ∞ ∞ X X X lim ak,n = lim ak,n = an. (†) k→∞ k→∞ n=1 n=1 n=1

1. Let ak,n = 1 when k = n and ak,n = 0 otherwise. Show that

∞ ∞ X X lim ak,n = 1 and an = 0. k→∞ n=1 n=1

2. (Monotone convergence theorem) Suppose (ak,n) consists of non-negative terms and that N X for each fixed n, the sequence (ak,n)k is increasing. Let Sk,N = ak,n. Show that n=1 0 0 Sk,N ≤ Sk0,N 0 whenever k ≤ k and N ≤ N , and hence show that

    lim lim Sk,N = lim lim Sk,N . k→∞ N→∞ N→∞ k→∞

Deduce that (†) holds.

As an application of this result, we consider the well-known constant e.

3. Show that k ∞  1  X k 1 1 + = , k n kn n=0 k
where n = 0 whenever n > k. 4. Show that n−1 k 1 1 Y  j  = 1 − n kn n! k j=0

k
n and deduce that n /k increases in k with limit 1/n! as k → ∞. 5. Apply the monotone convergence theorem to show that

k ∞  1  X 1 lim 1 + = . k→∞ k n! n=0 The common value is denoted e and is approximately 2.718.

6. Show that e is irrational by supposing for the sake of contradiction that e = p/q for p, q ∈ N, ∞ X 1 then showing that q!e = integer + = not an integer. (q + 1)(q + 2) ··· (q + j) j=1

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