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.

  • $\begingroup$ Apparently true for $s=-1$ as well: math.stackexchange.com/questions/599999 $\endgroup$ – punctured dusk Jan 1 '18 at 11:09
  • 6
    $\begingroup$ 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

It converges by Hardy's inequality: $$\sum_{n=1}^{\infty} \left(\frac{a_1^{1/s}+a_2^{1/s}+\cdots +a_n^{1/s}}{n}\right)^s\leq \left(\frac{s}{s-1}\right)^s\sum_{n=1}^{\infty} a_n.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.