Prove the series $\sum_{n=1}^{\infty} \frac{n^2{a_n}}{(a_1 + a_2 + ... + a_n)^{2}}$ converges

Question

Prove the series $\sum_{n=1}^{\infty} \frac{n^2{a_n}}{(a_1 + a_2 + \dots + a_n)^{2}}$ converges given $\sum_{n=1}^{\infty} \frac{1}{a_n}$ convereges, $\forall a_n > 0$

This is a question that a friend of mine has studied throughout the last semester and failed to solve. I think he would like to figure out this question but needs some hints.

Answered e.g. here: https://math.stackexchange.com/q/200514/42969 and here: https://math.stackexchange.com/q/2543985/42969 – both found with Approach0 — Martin R, Aug 05 '21 at 06:47

Wang YeFei · Answer 1 · 2021-08-05T05:19:04.860

2

Define $B_n = \dfrac{n^2a_n}{S^2_n}, S_n =\sum_{k=1}^n a_k, R_n= \sum_{k=1}^n \dfrac{1}{a_k}$. By Cauchy inequality twice: $S_n \ge \dfrac{n^2}{R_n}$. So $R_n \ge \dfrac{n^2}{S_n}$. Thus $B_n = \dfrac{n^2}{S_n}\cdot \dfrac{a_n}{S_n}\le \dfrac{a_n}{S_n}\cdot R_n\le R_n$. Thus by comparison test, $\sum B_n$ converges !

edited Aug 05 '21 at 05:19

answered Aug 05 '21 at 05:08

Wang YeFei

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How do you get $B_n \le \dfrac{a_n}{S_n}\cdot R_n$? – Kavi Rama Murthy Aug 05 '21 at 05:13
Incorrect argument. – Sungjin Kim Aug 05 '21 at 08:12
In fact $B_n$ should be $\le 25R_n$ – Il Prete Rosso Aug 05 '21 at 11:39
$B_n\leq R_n$ and $\lim_{n\rightarrow\infty} R_n<\infty$ are true. But, $\sum R_n$ may not converge. Thus, your argument by comparison test is incorrect. – Sungjin Kim Aug 05 '21 at 20:02

Prove the series $\sum_{n=1}^{\infty} \frac{n^2{a_n}}{(a_1 + a_2 + ... + a_n)^{2}}$ converges

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