Assume that $a_n>0$ such that $\sum_{n=1}^{\infty}a_n $ converges.
Question: For what values of $s\in \Bbb R$ does the following series : $$ I_s= \sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s.$$ converges or diverges?
This question is partially motivated by some comments on this post where it is shown that $I_s$ converges for $s>1$. Moreover, it is well known that $$\lim_{s\to\infty}\left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s = \left(a_1a_2\cdots a_n\right)^{1/n}$$
Accordingly, Taking $b_n= 1/a_n$ is one readily get,
$$\lim_{\color{red}{s\to-\infty}}\left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s = \left(a_1a_2\cdots a_n\right)^{1/n}$$ it draws from Carleman's inequality that :
$$\color{red}{ I_{-\infty}}=I_\infty= \sum_{n=1}^{\infty}\left(a_1a_2\cdots a_n\right)^{1/n} \le e \sum_{n=1}^{\infty} a_n<\infty .$$ Patently it is also true that the convergence holds for $s=-1$ this is proven here. Whereas the convergence fails for $0<s<1$ Indeed, $$\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s \ge \sum_{n=1}^{\infty} \frac{a_1}{n^s}=\infty$$
So we have that $I_s$ converges for $1<s\le\infty$ or $s=\in\{-1,-\infty\}$ and diverges for $0<s<1$.
Hence the original question reduces on studying $I_s$ for $s\le0$ can anyone help?
Clearly the hope is that $I_s$ converges for for $-\infty\le s\le -1 $ and diverges for $-1<s<0.$`
I don't know if one could infer some conjecture for the case $s=0$ since it seems pathological.