If $\sum a_n$ converges, so does $\sum a_n^{\frac{n}{n+1}}$ Let $a_n$ be a positive real sequence such that the series $\sum a_n$ converges.
I was asked to prove that under such circumstances $\sum a_n^{\frac{n}{n+1}}$ converges.
The previous sum can be rewritten as $\sum \frac{a_n}{a_n^{1/(n+1)}}$.
I can't prove the convergence of this last series... I'm sure comparison test is enough here, but other than $a_n \rightarrow 0$ there is no information that helps.
Thanks for your help.

Hint: if $a_n \leq 2^{-n-1}$, then $a_n^{n/(n+1)} \leq 2^{-n}$. Otherwise, $a_n^{n/(n+1)} \leq 2a_n$.
