An interesting exercise about converging positive series, involving $\sum_{n\geq 1}a_n^{\frac{n-1}{n}}$

Define $$S=\{n | a_n \leq \frac{1}{2^n}\}$$ $$T=\{n | \frac{1}{2^n} < a_n\}$$

Since $a_n$ is positive it suffices to show that $$\sum\limits_{n\in S} \frac{a_n}{\sqrt[n]{a_n}} \ \text{and} \ \ \sum\limits_{n\in T} \frac{a_n}{\sqrt[n]{a_n}} $$ converge separately.

If $n\in S$ then $$a_n^{\frac{n-1}{n}} \leq \frac{1}{2^{n-1}}$$ thus the first series converges.

If $n \in T$ then $$\frac{1}{2}<\sqrt[n]{a_n}$$ thus $$\frac{a_n}{\sqrt[n]{a_n}}<2a_n$$

ad the second series will also converge by comparison with $\sum\limits_{n\in T}2a_n$

My friend came up with the following solution. The proof is very similar to OP's one. It is a bit easier as it does not involve facts about harmonic numbers, logarithms and $e$; however the bound provided by OP is tighter.

AM-GM yields $$a_n^{(n-1)/n} = \sqrt[n]{\frac 1n \cdot na_n \cdot \underbrace{a_n \cdot \ldots \cdot a_n}_{n-2 \text{ times}}} < \frac 1n \cdot \left(\frac 1n + na_n + \underbrace{a_n + \ldots + a_n}_{n-2 \text{ times}}\right) < \frac {1}{n^2} + 2a_n.$$

Thus $$\sum_{n=1}^\infty a_n^{(n-1)/n} < \sum_{n=1}^\infty \frac 1{n^2} + 2\sum_{n=1}^\infty a_n < \infty.$$

Since $\sum_{n\ge 1}a_n$ is convergent, you can say that $a_n \approx_{n\to \infty} u_n$ with $u_n < {1\over n}$. Raising this inequality to to the power ${n-1}\over n$ you get :

$$ a_n^{{n-1}\over n}\approx_{n\to \infty} u_n^{{n-1}\over n} < {1 \over n^{1+{{n-1}\over n}}} = b_n $$

EDIT : The previous inequlity is wrong, here is the correct one : $$ a_n^{{n-1}\over n}\approx_{n\to \infty} u_n^{{n-1}\over n} < {1 \over n^{{{n-1}\over n}}} = b_n $$ Unfortunatly this does not provide enough with regard to Riemann rule to conclude to anything.

With $b_n$ convergent too by Riemann. Since all those sequences are positives, you can deduce $\sum_{n\ge 1}a_n^{{n-1}\over n}$ converge too.