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This question is a cross-post from MSE, cause I didn't get any answer there. I hope it is well suited for MO:

Let $(x_{n})_{n\ge 1}$ be an increasing sequence of positive integers and $\displaystyle{\overline{x}_{n}:=\dfrac{1}{n}\sum_{i=1}^{n}x_{i}}$. Suppose furthermore that $\forall\varepsilon\gt 0, \ \ n.\overline{x}_{n}\ll_{\varepsilon}n^{1+\varepsilon}$. Does it entail that $x_{n}\ll \overline{x}_{n}^{2}$?
Thanks in advance.

Edit: I think I must clear up my idea here. Writing $m_{n}:=\dfrac{x_{1}+x_{n}}{2}$ and using the inequality of arithmetic and geometric means, one gets $x_{1}.x_{n}\le m_{n}^{2}$. My idea is that, under the growth condition above, $m_{n}$ is roughly of the same order of magnitude as $\overline{x}_{n}$.

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  • $\begingroup$ What does $\ll$ mean? $\endgroup$ Sep 23, 2014 at 16:14
  • $\begingroup$ Why there are number theory tags here? $\endgroup$
    – Asaf
    Sep 23, 2014 at 16:15
  • $\begingroup$ $x\ll y$ means the same thing as $x=O(y)$. I added the number theory tags as number theorists are rather familiar with this notation, and the terms of the sequence I consider are positive integers. $\endgroup$ Sep 23, 2014 at 16:24
  • $\begingroup$ I suppose $x_n$ is not strictly increasing $\endgroup$ Sep 23, 2014 at 16:43
  • $\begingroup$ Not necessarily, indeed. $\endgroup$ Sep 23, 2014 at 16:45

1 Answer 1

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No. Let $(y_k)$ be a rapidly increasing sequence of positive integers, and for $y_k\leq n<y_{k+1}$ put $x_n:=y_k$. Note that $(x_n)$ is an increasing sequence of positive integers satisfying $x_n\leq n$. In particular, $\overline{x}_{n}<n/2$. On the other hand, for $n:=y_k$ we have $x_n=y_k$, while $\overline{x}_{n}<1+y_{k-1}$. Hence $x_{n}\ll \overline{x}_{n}^{2}$ would imply that $y_k\ll y_{k-1}^2$ which is not always true. For a concrete counterexample take $y_k:=2^{3^k}$.

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  • $\begingroup$ Thank you but does your proposed counterexample meet the requirement $n.\overline{x}_{n}\ll_{\varepsilon} n^{1+\varepsilon}$ for all $\varepsilon\gt 0$? It doesn't look obvious to me. $\endgroup$ Sep 23, 2014 at 16:56
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    $\begingroup$ @SylvainJULIEN: My example satisfies the stronger bound $x_n\leq n$ which clearly implies $\overline{x}_{n}<n/2$. $\endgroup$
    – GH from MO
    Sep 23, 2014 at 16:56

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