1

Let $a_n > 0$ and let $s_n = a_1 +\dots+ a_n$. Prove

(a) if $\sum a_n$ converges then $\sum \dfrac{ a_n}{s_n}$ converges.

(b) if $\sum a_n$ diverges then $\sum \dfrac{ a_n}{s_n}$ diverges but $\sum \dfrac{ a_n}{s_n^2}$ converges.

Trial: $\sum a_n$ converges to $l$ $\implies |\sum a_n -l|<\epsilon\,\, \forall n \ge N$ and with help of this I need to show $|\sum \dfrac{ a_n}{s_n}-l'|<\epsilon\,\, \forall n \ge N'$. But here I am stuck. Please help.

Argha
  • 4,671
  • For $(a)$, you need only show that the sum is finite, since $a_n > 0$. – Daniel Fischer Sep 01 '13 at 18:08
  • For the second one: show that $$\frac{a_{N+1}}{s_{N+1}} + \cdots \frac{a_{N+k}}{s_{N+k}} \ge 1 - \frac{s_N}{s_{N+k}},$$ and also that $$\frac{a_N}{s_N^2} \le \frac{1}{s_{N-1}} - \frac{1}{s_N}.$$ – Chris Sep 01 '13 at 18:12
  • https://math.stackexchange.com/questions/2961588/proving-that-sum-limits-n-1%e2%88%9e-fraca-ns-n2-converges, https://math.stackexchange.com/questions/808964/divergence-of-sum-limits-k-1n-fraca-ns-n-where-s-n-sum-limits-k – StubbornAtom Dec 07 '19 at 19:32

2 Answers2

2

For (a), note that $\frac{a_n}{s_n}\le \frac{a_n}{a_1}$, and use Comparison.

André Nicolas
  • 507,029
1

Fix $\epsilon >0$. For the first one, assuming $s_n \to l$ as $n \to \infty$, justify the claims that

$$(1): \text{There exists } N_1 : n> N_1 \implies l - \frac{l}{2} < s_n \le l,$$

which implies that $\sum_{n=N}^{M} \frac{a_n}{s_n} < \frac{2}{l}\sum_{n=N}^{M} a_n$ whenever $N > N_1$; and

$$(2): \text{There exists } N_2: N > N_2 \implies \sum_{n=N}^M a_n < \frac{l}{2}\epsilon$$

for any $M > N$. Use these two facts to show that $\sum_{n=N}^{M} \frac{a_n}{s_n} < \epsilon$ if $N > \max(N_1, N_2)$, and deduce that $\sum\frac{a_n}{s_n}$ converges.

I gave you some hints above for the second problem; see if you can use those.

Chris
  • 4,865