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Suppose that $\sum_{n=1}^{\infty}$ $a_n$ converges with each $a_n>0$, and let $s_n$ = $\sum_{k=1}^{n}$ $a_k$.

How to show that $\sum_{n=1}^{\infty}$ $a_n s_n$ and $\sum_{n=1}^{\infty}$ $\frac {a_n} {s_n}$ both converge.

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    $\begingroup$ Use the fact that $0<a<s_n<b$ along with the positivity of $a_n$ $\endgroup$ – Alex R. May 7 '15 at 19:19
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Prove the following lemma first:

If $b_n$ is a bounded sequence and $\sum a_n$ converges absolutely, then $\sum a_nb_n$ converges absolutely (hint: use the comparison test by saying that if $\sup_n |b_n| <M$, then $|a_nb_n| \leq |a_n|M$ for all $n$).

Then show that $s_n$ and $\frac{1}{s_n}$ are bounded sequences (since they both converge), and apply the preceding statement.

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Since $\sum_{n=1}^\infty a_n$ is absolute convergent and $a_n>0$ for each $n$ it holds $0 < a_1 \leq |s_n| \leq C$ for some $C>0$. It follows that $|\frac{1}{s_n}|\leq \frac{1}{a_1}$. Therefore $|a_ns_n| \leq |a_n|C$ and $|\frac{a_n}{s_n}|\leq a_na_1$ and by comparison test bost series converge because $\sum_{n=1}^\infty a_1a_n$ and $\sum_{n=1}^nCa_n$ both converge.

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Let $a=\sum_{n=1}^{\infty}a_n\,\,$, then $\lim_{n\to \infty}s_n=a\,$. Now consider that the series $\sum_{n=1}^{\infty}aa_n$ is convergence. Let: $b_n=a_ns_n$ then, $\lim_{n\to \infty}\frac{b_n}{aa_n}=\lim_{n\to \infty}\frac{a_ns_n}{aa_n}=\lim_{n\to \infty}\frac{s_n}{a}=1$. So $\sum_{i=1}^\infty b_n=\sum_{i=1}^\infty a_ns_n\lt\infty$. similarly, for $\sum_{i=1}^\infty\frac{a_n}{s_n}$, consider the series $\sum_{i=1}^\infty \frac{a_n}{a}$

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