Yesterday I stumbled across an interesting exercise (Indam test 2014, Exercise B3):

(Ex)Given a positive sequence $\{a_n\}_{n\geq 1}$ such that $\sum_{n\geq 1}a_n$ is convergent, prove that $$ \sum_{n\geq 1}a_n^{\frac{n-1}{n}}$$ is convergent, too.

My proof exploits an idea from Carleman's inequality. We have: $$ a_n^{\frac{n-1}{n}}=\text{GM}\left(\frac{1}{n},2a_n,\frac{3}{2}a_n,\ldots,\frac{n}{n-1}a_n\right) $$ and by the AM-GM inequality $$ a_n^{\frac{n-1}{n}}\leq \frac{1}{n}\left(\frac{1}{n}+a_n\sum_{k=1}^{n-1}\frac{k+1}{k}\right)\leq \frac{1}{n^2}+\left(1+\frac{\log n}{n}\right)a_n $$ hence $$ \sum_{n\geq 1}a_n^{\frac{n-1}{n}}\color{red}{\leq} \frac{\pi^2}{6}+\left(1+\frac{1}{e}\right)\sum_{n\geq 1}a_n.$$

Now my actual

Question:Is there a simpler proof of(Ex), maybe through Holder's inequality, maybe exploiting the approximations $$ \sum_{m<n\leq 2m}a_n^{\frac{2m-1}{2m}}\approx \sum_{m<n\leq 2m}a_n^{\frac{n-1}{n}}\approx \sum_{m<n\leq 2m}a_n^{\frac{m-1}{m}}$$ "blocking" the exponents over small summation sub-ranges?