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Assume that $a_n\ge0$ such that $\sum_{n=1}^{\infty}a_n $ converges, then

show that for every $s>1$ the following series converges too: $$\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s.$$

I failed to handle this with Hölder inequality. Any tips or hint will be appreciated.

Also it might be helpful to see that there is a Césaro sum of $(a_n^{1/s})_n$ appearing in the last series.

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  • $\begingroup$ Apparently true for $s=-1$ as well: math.stackexchange.com/questions/599999 $\endgroup$ – punctured dusk Jan 1 '18 at 11:09
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    $\begingroup$ You seemed well-aware of Hardy's inequality here, it is a strange question coming from you. $\endgroup$ – Jack D'Aurizio Jan 1 '18 at 11:30
  • $\begingroup$ @barto patently there must be some gap between $0$ and $1$ : I think one should be able to prove that this failed for $0<s<1. $ $\endgroup$ – Guy Fsone Jan 1 '18 at 11:38
  • $\begingroup$ @GuyFsone Indeed, already for $(a_n)=(1,0,0,\ldots)$ $\endgroup$ – punctured dusk Jan 1 '18 at 12:05
  • $\begingroup$ @barto that wise now can we shift backward a bitt to $-1<s<0$? from this we have more clue for the range of $s$ $\endgroup$ – Guy Fsone Jan 1 '18 at 12:10
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It converges by Hardy's inequality: $$\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s\leq \left(\frac{s}{s-1}\right)^s\sum_{n=1}^{\infty} a_n.$$

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