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I need some help here. I started with $$\sum_{k=1}^N \frac{1}{k^2 H_k}=\sum_{k=1}^N \frac{1}{k^2 \int_0^1\frac{1-x^k}{1-x}dx}=\sum_{k=1}^N \frac{1}{\int_0^1 k^2 \frac{1-x^k}{1-x}dx}$$

But I do not know how to continue anymore.

Cave Johnson
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kylexy
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    Given that the sum converges, I'm not sure what "asymptotic estimate" would mean, apart from computing the limit... Anyway, do you have any reason to suppose that this has some simple solution? – leonbloy Nov 05 '16 at 14:50

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Hint. Note that Harmonic numbers have a nice asymptotic estimate $$H_n\approx\ln n+\gamma+o(1)$$ where $\gamma\approx 0.5772156649$ is the Euler-Mascheroni constant. Moreover for $n\geq 1$, $$\frac{1}{2}<H_n-\ln n\leq 1.$$

Robert Z
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  • am I right in thinking that it follows that $\Sigma_{k=1}^N \frac{1}{k^2 * H_k} < \Sigma_{k=1}^N \frac{2}{(1+2 \ln k )k^2 }$? Am I too stupid to have this thinking? I am really lost. – kylexy Nov 05 '16 at 13:04
  • @kylexy Your inequality is fine. Since the series $sum_{k=1}^{\infty}\frac{1}{k^2 H_k}$ is convergent, you can estimate the sum. What are you exactly looking for? – Robert Z Nov 05 '16 at 13:13
  • I am looking for the asymptotic estimate, so I figured that I should look for the value of the finite sum... I don't know if i am on the right track. – kylexy Nov 05 '16 at 13:16
  • Am I right to conclude that $\Sigma_{k=1}^N \frac{1}{k^2 H_k}$ is convergent because the $$lim_{k\rightarrow \infty} \frac{2}{(1+2 \ln k) k^2 }=0$$? – kylexy Nov 05 '16 at 13:19
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    @kylexy No, it is convergent because $1/k^2/H_k\leq 1/k^2$ and $\sum 1/k^2$ is convergent. – Robert Z Nov 05 '16 at 13:25
  • @Robert Z Am I right to conclude that the asymptotic estimate to this problem is O(1/k^2)? – desperatemuch Nov 29 '16 at 00:28
  • @desperatemuch The asymptotic estimate of $1/(k^2H_k) $ is $O(1/k^2/\ln(k))$ (or $O(1/k^2)$). The series is convergent. – Robert Z Nov 29 '16 at 08:02