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.

We have that by AM-GM

$$a_n^{\frac{n}{n+1}}\le \frac{2a_n+2a_n+\ldots+2a_n+\frac1{2^n}}{n+1}=\frac{2na_n+\frac1{2^n}}{n+1} \le2a_n+\frac1{2^n}$$

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 $$ \sum_{n=1}^\infty a^{n/(n+1)}_n\le \frac{1}{2}+2A<\infty. $$

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