For all n ≥1 , set

2.3.7. For all n ≥1 , set

s n = ∑ a k and σ n = ∑ 1 −

n +1 Assume that the series ∞ ∑

k =1

k =1

n =1 |s n − σ n | converges, for some α >0 . Prove that the series ∑ n =1 a n is convergent.

Solution. Since s n − σ n → 0 as n→∞, it is sufficient to prove that the sequence ( σ n ) n ≥1 converges. We have

n − σ n −1 =

∑ ka k =

n (n + 1) k =1

76 2 Series Summing for n = 1, 2, . . . , N, we obtain

n =1

If α ≤ 1, the convergence is immediate. If α > 1, H¨older’s inequality yields

|s n − σ n |

≤ ∑ |s n − σ α

where 1 / α + 1/ β = 1. Since both factors on the right-hand side converge to 0, so does the left-hand side. Thus, the sequence ( σ n ) n ≥1 is Cauchy, hence convergent. ⊓ ⊔

2.3.8. Let (a n ) n ≥1

be a sequence of real numbers such that the series ∑ ∞ n =1 a 4 n is

convergent. Prove that the series ∑ ∞ n =1 a 5 n is convergent.

Solution. The hypothesis implies that a 4 n →0 as n→∞. Thus, there exists a posi- tive integer N such that a 4 5 n 4 < 1 for all n ≥ N; hence |a

5 n |<a n , provided n by the first comparison test, the series ≥ N. So, ∑ n =1 a n is convergent. ⊓ ⊔

2.3.9. Let (a n ) n ≥1

be a sequence of positive numbers such that a n and the series ∞ ∑ n =1 a n is convergent. Prove that the series ∞ ∑ n =1 a n /(1 − a n ) is convergent.

Solution. An argument similar to that developed in the previous exercise imp- lies that the series ∑ ∞

n =1 a 2 n is convergent. Since the series ∑ n =1 a n converges, it follows that a n →0 as n→∞ and there exists an integer N such that 1 − a n > 0 and

a n /(1 − a n ) < 2a n + 2a 2 n for all n ≥ N. Thus, by the first comparison test, the series ∑ ∞ n =1 a n /(1 − a n ) is convergent. ⊓ ⊔

2.3.10. Let (a n ) n ≥1

be a decreasing sequence of positive numbers. Prove that if there is a positive number k such that a n ≥ k/n for infinitely many n , then the series

∑ ∞ n =1 a n is divergent. Solution. Fix ε > 0 such that ε ≤ k/3. For N arbitrarily large, select n > 3N such that a n ≥ k/n. Let m be a positive integer such that N < m < 2N. Thus, m/n < 2/3. We have

3 So, in view of Cauchy’s criterion, the series ∞ ∑ n =1 a n is divergent. ⊓ ⊔

2.3.11. Let (a n ) n ≥1

be a sequence of nonnegative numbers such that a 2n −

a 2n +1 ≤a 2 n , a 2n +1 −a 2n +2 ≤a n a n +1 for any n ≥1 and lim sup n → ∞ na n < 1/4 . Prove that the series ∑ ∞ n =1 a n is convergent.

n → ∞ a n < 1. Then, by the root test, we obtain the conclusion. Let c j = sup

Solution. We prove that lim sup

n ≥2 j (n + 1)a n for j ≥ 1. We show that c j +1 ≤ 4c j . Indeed, for any integer n ≥2 j +1 there exists an integer k ≥2 j such that either n = 2k or n = 2k + 1.

2.3 Convergent and Divergent Series 77 In the first case we have

whereas in the second case we obtain

4c j 4c j 2k +1 −a 2k +2 ≤a k a k +1 ≤ (k + 1)(k + 2) ≤ 2k +2 −

. 2k +3

It follows that the sequence (a n −4c 2 j /n + 1) n ≥2 j +1 is nondecreasing and its terms are nonpositive, since it converges to 0. Therefore a

2 j n +1 ≤ 4c j /(n + 1) for n ≥ 2 , that is, c 2 2 j −j +1 ≤ 4c j . This implies that the sequence ((4c 2 j ) ) j ≥0 is nonincreasing and therefore bounded from above by some number q

∈ (0,1), since all its terms except finitely many are less than 1. Hence c j ≤q 2 for l large enough. For any n between 2 j and 2 j +1 we have

≤q ≤( q ) n .

n +1 Hence lim sup √ n a n √ ≤ q < 1, which concludes the proof. ⊓ ⊔