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Let $\{a_n\}_{n=1}^{\infty}$ be a sequence of positive numbers and let $b_{n} = \frac{a_{n}}{(a_{1}+...+a_{n})^{2}}$ for n $\in\mathbb{N}$. Prove that $\sum_{n=1}^{\infty}b_{n}$ is a convergent series. I'm stuck on how to start this problem. I've considered the limit comparison test, but it hasn't worked out for me. I know I can assume $\{a_{n}\}$ and $\{b_{n}\}$ are positive, so maybe I need to show $\{b_{n}\}$ has an upper bound and apply the positive series test. Any help is appreciated. Thank you!

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Hint:

With $S_n = a_1+ \ldots + a_n$,

$$\frac{a_n}{S_n^2} \leqslant \frac{S_n - S_{n-1}}{S_nS_{n-1}} = \frac{1}{S_{n-1}}- \frac{1}{S_n}$$

Note that $S_n$ converges either to a finite limit or $+\infty$ and in all cases $1/S_n $ converges to a finite limit.

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  • $\begingroup$ Clever approach and definitely makes sense to me. Thank you! $\endgroup$ – dorkichar Jan 3 at 21:24

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