# For $a_n>0$ such that $\sum a_n$ converges, show that there exist $c_n>0$ such that $c_n\to \infty$ and $\sum a_n c_n$ is finite.

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.

• See here for a more general problem. – David Mitra Aug 25 '13 at 17:55

$$r_n=\sum_{k=n+1}^\infty a_n$$ 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 $c_n=\frac{1}{\sqrt{r_{n-1}}}$.
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.