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.

  • 3
    $\begingroup$ See here for a more general problem. $\endgroup$ – 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 $c_n=\frac{1}{\sqrt{r_{n-1}}}$.

  • 2
    $\begingroup$ Neat one, Sami. $\endgroup$ – Pedro Tamaroff Aug 25 '13 at 18:09

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.