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. ⊓ ⊔