I want to prove that if $\sum_{n=1}^\infty a_{n}$ converges and $a_n \ge 0$ $\forall n \in \mathbb{N}$, then $\sum_{n=1}^\infty a_{n}^{1-\frac{1}{n}}$ converges as well.
This is what I've done so far: If $\sum_{n=1}^\infty a_{n} \Rightarrow$ $\lim_{n \to \infty}a_n=0$.
$\lim_{n \to \infty} \ln(\frac{a_n}{\sqrt[n]{a_n}})=\lim_{n \to \infty} (1-\frac{1}{n})\ln(a_n)=-\infty \Rightarrow \lim_{n \to \infty} \frac{a_n}{\sqrt[n]{a_n}}=0$ and so it is possible that $\sum_{n=1}^\infty a_{n}^{1-\frac{1}{n}}$ converges.
$\lim_{n \to \infty} \ln(\sqrt[n]{a_n})=\lim_{n \to \infty} \frac{1}{n}\ln(a_n)=0 \Rightarrow \lim_{n \to \infty} \sqrt[n]{a_n}=1$.
Let $s_k=\sum_{n=1}^k a_{n}^{1-\frac{1}{n}}$, $s_{k+1}-s_k=a_{k+1}^{1-\frac{1}{k+1}} >0$ $\Rightarrow s_k$ is increasing.
Since $a_n \ge 0$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty}a_n=0$, there must be $M \in \mathbb{N}$ so that $a_M\ge a_n, \forall n \in \mathbb{N}$
$s_k=\sum_{n=1}^k a_{n}^{1-\frac{1}{n}}\le a_M\sum_{n=1}^k \frac{1}{\sqrt[n]{a_n}}$.
If I show that $\sum_{n=1}^\infty \frac{1}{\sqrt[n]{a_n}}=L$ converges then $s_k\le a_M\sum_{n=1}^k \frac{1}{\sqrt[n]{a_n}}<a_M\sum_{n=1}^\infty \frac{1}{\sqrt[n]{a_n}}=a_ML$ $\Rightarrow s_k$ is increasing and bounded so converges and $\sum_{n=1}^\infty a_{n}^{1-\frac{1}{n}}$ converges as well.
I'm finding troubles in proving that $\sum_{n=1}^\infty \frac{1}{\sqrt[n]{a_n}}$ converges.
Thanks for help.
