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Does there exist a sequence $a_n\in \mathbb{R}^{>0}$ such that $$\liminf_{N\rightarrow \infty}~~~ \frac{1}{N}\sum_{n=1}^Na_n >0$$

and $$\sum_{n=1}^\infty \frac{1}{a_n\cdot n^2} = \infty~~?$$

P.S. I posted the logical next question here Does there exist a non-increasing sequence with these properties?

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  • $\begingroup$ The second condition is impossible for $a_n\in \Bbb N$ because then $0<1/(a_nn^2)\leq 1/n^2 $ and $\sum_{n=1}^{\infty}(1/n^2)<\infty.$ $\endgroup$ Jan 28 '18 at 5:57
  • $\begingroup$ What about $a_n=\frac{\ln n}{n}$ $\endgroup$ Jan 28 '18 at 5:57
  • $\begingroup$ @GevorgHmayakyan, Then $\sum 1/(n^2a_n)$ converges. $\endgroup$ Jan 28 '18 at 5:59
  • $\begingroup$ Sorry edited just now $\endgroup$ Jan 28 '18 at 5:59
  • $\begingroup$ @DanielWainfleet yes, sorry about that. $\endgroup$
    – user525678
    Jan 28 '18 at 6:04
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Let $a_n= 1/n^2$ when $n$ is even, and $1$ when $n$ is odd. Then the average converges to $1/2$ and the sum is $\infty$.

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  • $\begingroup$ The OP asks for a sequence of natural numbers, though. $\endgroup$ Jan 28 '18 at 6:00
  • $\begingroup$ @GTonyJacobs Oops. Gotta read all the words. $\endgroup$
    – saulspatz
    Jan 28 '18 at 6:02
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    $\begingroup$ @GTonyJacobs sorry. my mistake. the second condition doesn't make sense for natural numbers. $\endgroup$
    – user525678
    Jan 28 '18 at 6:02
  • $\begingroup$ What I see in the Q is $\sum (a_n n^2)^{-1}.$ For $a_n\in \Bbb N$ we have $0<(a_nn^2)^{-1}\leq n^{-2}$ so the sum must be finite because $\sum_{n=1}^{\infty}n^{-2}=\pi^2 /6$. $\endgroup$ Jan 28 '18 at 6:04
  • $\begingroup$ @DanielWainfleet When $n$ is even, $a_nn^2 = 1,$ so the summand is $1$ infintely often. $\endgroup$
    – saulspatz
    Jan 28 '18 at 6:07

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