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A problem From PhD Pre lims Exam:

Let $ a_{n} > 0 $ for all $ n\in\mathbb{N}$ such that $\sum a_{n} $ converges. Show that there exist $ c_{n} > 0 $ ($n\in\mathbb{N}$) such that $ \lim \limits_{n\to\infty} c_{n}= \infty $ and $\sum a_{n}c_{n} $ is finite.

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    $\begingroup$ See here for a more general problem. $\endgroup$ – David Mitra Aug 25 '13 at 17:55
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Let

$$r_n=\sum_{k=n+1}^\infty a_n$$ so since the series $\sum a_n$ is convergent then the sequence $(r_n)$ is decreasing and convergent to $0$.

We have $$\frac{a_n}{\sqrt{r_{n-1}}}=\frac{r_{n-1}-r_n}{\sqrt{r_{n-1}}}=\frac{(\sqrt{r_{n-1}}-\sqrt{r_{n}})(\sqrt{r_{n-1}}+\sqrt{r_{n}})}{\sqrt{r_{n-1}}}\leq2(\sqrt{r_{n-1}}-\sqrt{r_{n}})=t_n$$ and since the series $\sum t_n$ is convergent (telescoping series) then the series $$\sum \frac{a_n}{\sqrt{r_{n-1}}}$$ is convergent. Take $c_n=\frac{1}{\sqrt{r_{n-1}}}$.

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    $\begingroup$ Neat one, Sami. $\endgroup$ – Pedro Tamaroff Aug 25 '13 at 18:09
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For each natural number $k$, find $N_k$ such that $\sum_{n=N_k}^\infty a_n < 1/4^k$. (and make sure $N_1 < N_2 < \ldots$.) Then set $c_n = 2^k$ for all $n$ in the range $N_k \leq n < N_{k+1}$. You will find that $\sum_{n=N_k}^\infty c_n a_n < 1/2^{k-1}$ by using geometric series.

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