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 $A=\sum_{n=1}^\infty a_n$, and
set $$S=\big\{n: a_n^{n/(n+1)}\le 2a_n\big\},$$ and $$T=\big\{n: a_n^{n/(n+1)}> 2a_n\big\}.$$ 
If $n\in T$, then 
$$
a^{n/(n+1)}_n> 2a_n\quad\Longrightarrow\quad a_n^n>2^{n+1}a_n^{n+1}\quad\Longrightarrow\quad
a_n<2^{-n-1}.
$$
$$
\sum_{n=1}^\infty a^{n/(n+1)}_n=
\sum_{n\in S} a^{n/(n+1)}_n+\sum_{n\in T} a^{n/(n+1)}_n\le 2\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty 2^{-n-1}=2\sum_{n=1}^\infty a_n+\frac{1}{2}.
$$
Thus
$$
\sum_{n=1}^\infty a^{n/(n+1)}_n\le
\frac{1}{2}+2A<\infty.
$$
