Conditioning series with positive real numbers

Question

I have to prove that if $\sum_{n=1}^{\infty} a_{n}$ is a convergent series with positive real numbers, then $\sum_{n=1}^{\infty} (a_{n})^\frac{n}{n+1}$ converges. I also wonder if the converse is true. Any suggestion, hint will
be very welcome. Thanks.

Answer 1

if $a_n \geq \frac{1}{2^{n+1}} \Rightarrow a_n^{\frac{n}{n+1}} =\frac{a_n}{a_n^{\frac{1}{n+1}}} \leq 2a_n$ if $a_n \leq \frac{1}{2^{n+1}}$ then $a_n^{\frac{n}{n+1} } \leq \frac{1}{2^n}$. Therefore $a_n^{\frac{n}{n+1} } \leq 2a_n + \frac{1}{2^n}$ and by comparison test you are done.