Here are the approaches that I've tried so far:
If $a_n^{-1/(n+1)}$ is bounded, say $a_n^{-1/(n+1)} \le c,$ then $\sum\limits_{n \ge 1} a_n^{n/(n+1)} \le \sum\limits_{n \ge 1} ca_n,$ so the series converges by the comparison test. Else, $\lim\sup a_n^{-1/(n+1)} = \infty \, (*) \, \Rightarrow \lim\inf a_n^{1/(n+1)} = 0 \Rightarrow -\infty = \lim\inf \frac{\log(a_n)}{n+1}.$ But $\sum\limits_{n \ge 1} a_n $ converges, so $\lim\sup \frac{a_{n+1}}{a_n} \le 1 \Rightarrow 0 \ge \lim\sup \log\left(\frac{a_{n+1}}{a_n}\right) = \lim\sup \frac{\log(a_{n+1})-\log(a_n)}{(n+2)-(n+1)} = \lim\sup \frac{\Delta \log(a_n)}{\Delta (n+1)}.$ My idea was to use Stolz-Cesaro here to derive a contradiction, but because we have $\lim \sup$ in $(*)$ instead of just $\lim,$ we are forced to work with $\lim \sup, \lim \inf.$ To top it all off, we can't even get the infinums and supremums to be pointing in the right directions. I don't think that the convergence of $\sum\limits_{n \ge 1} a_n^{n/(n+1)}$ can be proved directly from $(*),$ and I'm not sure what other ideas we can use.
Another idea I had was to suppose $M = \sum\limits_{n=1}^{\infty} a_n$ and let $\sum\limits_{n=1}^{b_i} a_n \ge M(1-2^{-i})$ with $b_1, b_2, \dots$ minimal. Then $\sum\limits_{n=b_i+1}^{b_{i+1}} a_n < \frac{M}{2^i},$ and through some fiddling with the power mean inequality we get $\sum\limits_{b_i + 1}^{b_{i+1}} a_n^{n/(n+1)} < (b_{i+1}-b_i)^{1/(b_i+1)} (2^{-i}M)^{b_i/(b_i+1)}.$ But now we need to show $\sum\limits_{i \ge 1} (b_{i+1}-b_i)^{1/(b_i+1)} (2^{-i}M)^{b_i/(b_i+1)}$ converges given $\{b_i\}$ is monotonic increasing, which may not be true due to how much information about $\{b_i\}$ we threw out.
I noticed that we may assume $\{a_n\}$ is monotonic decreasing. Then we have a more manageable bound: $\sum\limits_{n=b_i+1}^{b_{i+1}} a_n \le (b_{i+1}-b_i)a_{b_i}^{b_i/(b_i+1)}.$ But this looks similar to the series we wish to show converges, so it appears we have not made any progress at all. We also have $(b_{i+1}-b_i)a_{b_{i+1}} < \frac{M}{2^i},$ so this sum is further bounded by $\frac{M}{2^i} a_{b_{i+1}}^{-1} a_{b_i}^{b_i/(1+b_i)} = \frac{M}{2^i} a_{b_{i+1}}^{-1/(1+b_i)} \left(\frac{a_{b_i}}{a_{b_{i+1}}}\right)^{b_i/(1+b_i)}.$ We've gotten rid of the pesky $b_{i+1}-b_i$ term and introduced the rapid decay inducing term $2^{-i},$ but now we have to deal with $a_{b_i}/a_{b_{i+1}}.$ But there's no guarantee that $a_{b_i}/a_{b_{i+1}}$ is well behaved, i.e. small. It could even be unbounded.
Yet another possible path: At most $M \cdot 2^{i+1}$ terms lie in the interval $[2^{-i-1}, 2^{-i}],$ so the contribution to the sum $a_n^{n/(n+1)}$ coming from $n$ such that $a_n \in [2^{-i-1}, 2^{-i}]$ could be estimated, if only we had a lower bound on $n.$ Without any lower bound on $n$ we can only conclude $a_n^{n/(n+1)} < a_n \le 2^{-i},$ which gives the estimate $M \cdot 2^{i+1} \cdot 2^{-i} = M/2.$ But $\sum\limits_{i \ge 1} M/2$ is divergent, so we need a stronger estimate.
Are any of these approaches on the right track? Do you have any hints or suggestions about how to finish the problem? I suppose the worst case scenario would be that none of my ideas lead anywhere and I have to tear down all this progress and start from scratch again. Hopefully this is not the case.
