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.

  • $\begingroup$ Hardy's inequality for negative exponents is also known as the Polya-Knopp inequality: just google it. The proofs exploit similar ideas (Holder+induction, or some form of convexity, or Bocharova's approach) $\endgroup$ Jan 1, 2018 at 18:50
  • $\begingroup$ You might have a look at the Cauchy-Schwarz and beyond section of my notes, too. $\endgroup$ Jan 1, 2018 at 18:52
  • $\begingroup$ @JackD'Aurizio Seriously I feel stupid after your comment. oooo god please step by step. I only know Hardy holder and convexity where do the rest come from?` $\endgroup$
    – Guy Fsone
    Jan 1, 2018 at 18:53

1 Answer 1


Answer: Using the power mean (generalized mean) inequality the convergence for $s<0$ follows easily from the convergence for $s>1$. This probably does not give optimal bounds for $I_s$.

It's more natural to set $s=1/t$, so that the integrands are generalized means that, for fixed $(a_n)$, are increasing in $t \in \mathbb R$ by the power mean inequality (which says exactly that). We know that $I_{1/t}$ converges for $0<t<1$ (i.e. $s>1$, by Hardy's inequality, see here) and thus for all $-\infty \leq t < 1$ by the comparison test.

We also have that:

  • the supremum $S_t=\sup(I_{1/t}/\sum a_n)$ over all sequences is finite for all $t<1$
  • $S_t$ is (monotonically) increasing in $t$
  • $S_{-\infty}=1$ (take a decreasing sequence)
  • $S_{-1}=2$ (see this question)
  • $S_0 \leq e$ (Carleman's inequality)
  • $S_t \leq (1-t)^{-1/t}$ for $0<t<1$ (Hardy's inequality, see this question)

It's natural to conjecture that $S_t=(1-t)^{-1/t}$ for all $t<1$, which is $1$ at $-\infty$ and $e$ at $0$.

  • $\begingroup$ there is one conclusion in your first statement that I don't understand. why is this converges for t less than one ? $\endgroup$
    – user503348
    Jan 2, 2018 at 6:36
  • $\begingroup$ I think I have don't what the Op did already check the question again $\endgroup$
    – user503348
    Jan 2, 2018 at 6:39
  • $\begingroup$ this answer is not clear $\endgroup$
    – user503348
    Jan 2, 2018 at 6:39
  • $\begingroup$ @Sobolev I was using a result from a previous question of the OP, see edit. $\endgroup$ Jan 2, 2018 at 9:52
  • $\begingroup$ @barto Sorry are you not to just repeating what I have say in the question?n the crucial point is to find the convergence for $ s<0$ which I cannot figure out in your answer $\endgroup$
    – Guy Fsone
    Jan 2, 2018 at 11:41

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