Suppose that (a n ) n ≥1 is a sequence of real numbers such that

2.3.14. Suppose that (a n ) n ≥1 is a sequence of real numbers such that

a n = ∑ a 2 n +k for n = 1, 2, . . . .

k =1

Show that if the series ∞ ∑ n =1 a n converges, then a n =0 for all n . Solution. We first observe that 0

≤a ∞ n +1 ≤a n for all n ≥ 1. If ∑ j =1 a j converges, we take k

≥ 1 such that ∑ ∞

j =k+1 a j < 1. Then

a k +1 ≤a k =

j ≤a k +1

∑ a j ≤a k +1 .

j =k+1

j =k+1

k =a k +1 and so a k +1 = 0, implying a j = 0 for j > k since a k +1 =∑ j =k+1 a 2 j . By induction we deduce that a j = 0 for all j < k + 1. ⊓ ⊔

Hence a

2.3.15. Let (a n ) n ≥1

be an increasing sequence of positive numbers such that

a n →∞ as n →∞ . Then ∞ ∑ n =1 (a n +1 −a n )/a n +1 diverges. Solution. We first observe that

a n −a ∑ 1

Pick n 1 so large that (a n 1 −a 1 )/a n 1 > 1/2. In general, given n r , pick n r +1 so large that (a n r +1 −a n r )/a n r +1 > 1/2, which is possible since (a n −a n r )/a n → 1 as n → ∞.

2.3 Convergent and Divergent Series 79 Now

∑ (a k +1 −a k )/a k +1 = ∑ (a k +1 −a k )/a k +1 + ∑ (a k +1 −a k )/a k +1 + ···

+ ∑ (a k +1 −a k )/a k +1

k =n r −1

≥ (a n 1 −a 1 )/a n 1 + (a n 2 −a n 1 )/a n 2 + ··· + (a n r −a n r −1 )/a n r > r/2 → ∞ as r → ∞.

Thus, since each term of ∞ ∑ k =1 (a k +1 −a k )/a k +1 is positive, the series diverges. The well known Raabe’s test for convergence or divergence of a positive series

∑ ∞ n =1 a n rests on knowing the behavior of an associated sequence

n : =n 1 − n

a +1

As established in Knopp [59], the series

n =1 a n converges, provided lim inf n → ∞ R n >1. If R n ≤ 1 for all sufficiently large n, then the series diverges.

A distinct associated sequence is

n > 1, then the series ∑ n =1 a n converges. If ˆ R n ≤ 1 for all sufficiently large n, then the series diverges. These two versions of Raabe’s test are not equivalent. Indeed, consider the series

If lim inf n → ∞ R ˆ

∞ (n − 1)!n!4 n

n =1

(2n)! n

Since

n − n 2 +o n 2 , we have

a n +1 2n

2n

as n →∞, which means that the test is inconclusive. On the other hand,

n =1+ 8 +o

. Therefore

n +1

1 + − 8 =1− 2 +o

a n 2n +1

n +1

n =1− +o n n 2 →1

as n →∞,

80 2 Series which means that the series diverges. However, removing the first term of the above

investigated series, we obtain ∞ (n − 1)!n!4 n

n ! (n + 1)! 4 n +1

n =1 (2n + 2)! n +1 For this last series we find that

n =2

(2n)! n

and hence

R ˆ n =1− 8 +o

→1 − as n →∞.

We conclude the divergence of the series by the second form of Raabe’s test.

As in [95], we introduce the following “parametrized” associated sequence

where k runs over all nonnegative integers. ⊓ ⊔

2.3.16. Let ∞ ∑ n =1 a n

be a series of positive terms.

(i) If R ˆ (k) n ≤1 for a nonnegative integer k and for all sufficiently large indices n , then the series diverges. (ii) If

a −1 =1+O

n +1

then the series diverges. Solution. (i) Indeed, the inequality ˆ R (k) n ≤ 1 implies

a n 1 +1 n +1−k

1 n −k

for all sufficiently large n. Thus since the series ∑ n >k 1 /n − k diverges, by the ratio comparison test the series ∑a n diverges as well.

(ii) The equality implies that

a n +1

for some positive integer M and for all sufficiently large n.

2.3 Convergent and Divergent Series 81 Thus

R ˆ (M+1)

and hence, by (i), the series ∞ ∑ n =1 a n diverges. ⊓ ⊔

2.3.17. Assume that a n >0 for each n , and that ∑ ∞ n =1 a n converges. Prove that the series ∞ ∑ n =1 a (n−1)/n n converges as well. Solution. Applying the AM–GM inequality, we have, for any n ≥ 2,

(n−1)/n

n (because 2xy ≤x +y ), and (n − 2)a n /n ≤ a n (because (n − 2)/n ≤ 1). Therefore, 0 <a (n−1)/n n ≤ 1/n 2 + 2a n , for each n ≥ 1. Finally, the comparison test

shows that ∞ a (n−1)/n converges, since ∑ ∞ n =1 n ∑ n =1 1 /n 2 +2∑ ∞ n =1 a n clearly converges. An alternative solution is based on the observation that each term a n satisfies either the inequality 0

≤ 1/2 n or 1 /2 n <a n . In the first case, a n ≤ 1/2 n −1 . In the second one,

<a (n−1)/n

a (n−1)/n

= 1 /n ≤ 2a n .

Therefore, in both cases,

<a n

0 (n−1)/n

≤ 2 n + 2a n .

The conclusion is now immediate since ∑ ∞ n =1 1 /2 n converges, and so does ∑ ∞ n =1 2a n . In the next exercise we give a full description of Bertrand’s series. ⊓ ⊔