If $\sum\limits_{n=1}^{\infty} a_n$ converges and $a_n > 0$ for all $n,$ then $\sum\limits_{n=1}^{\infty} a_n^{n/(n+1)}$ converges.
If $\sum_{n=1}^{\infty} a_n$ converges, then so does $\sum_{n=1}^{\infty} a_n^{\frac{n}{n+1}}$