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.

Let $n$ be a positive integer. 
If $a_n \le \frac{1}{2^{n+1}}$, then $a_n^{\frac{n}{n+1}}\le \frac{1}{2^n}$. 
If $a_n \gt  \frac{1}{2^{n+1}}$, then by your calculation $a_n^{\frac{n}{n+1}}\le 2a_n$.
Thus for all $n$ we have $a_n^{\frac{n}{n+1}}\le \frac{1}{2^n}+2a_n$. 
Let $b_n=\frac{1}{2^n}+2a_n$. Then $\sum b_n$ converges, and therefore by Comparison so does our sum.
