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I was trying to solve some problem and came across the following series: $$ \sum_{k=2}^{\infty}\frac{H_k}{k^2} $$

I tried to find a closed form for that series but could not. Also I looked some articles related to harmonic sum but it did not help me too.

Any hints and suggestions would be appreciated.

Thanks!

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  • $\begingroup$ Do you want the exact results or asymptotics? $\endgroup$ – Alex Feb 20 '15 at 14:26
  • $\begingroup$ You can use the fact that $H_k=\displaystyle \int_0^1 \frac{1-x^k}{1-x}\,dx$. $\endgroup$ – Tolaso Feb 20 '15 at 14:27
  • $\begingroup$ I want exact results(if they exist of course). @Tolaso I tried to use that identity, but after substituting it in the series could not go further. $\endgroup$ – pointer Feb 20 '15 at 14:32
  • $\begingroup$ It is equal to $-1+2\zeta(3)$. Where $\zeta(3)$ is transcendental. $\endgroup$ – Tom Feb 20 '15 at 14:35
  • $\begingroup$ Did you swip integration and summation? $\endgroup$ – Tolaso Feb 20 '15 at 14:36
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Following the suggestion I gave above: $$\begin{aligned} \sum_{k=1}^{\infty}\frac{\mathcal{H}_k}{k^2} &=\sum_{k=1}^{\infty}\frac{1}{k^2}\int_{0}^{1}\frac{1-x^k}{1-x}\,dx \\ &=\int_{0}^{1}\frac{1}{1-x} \sum_{k=1}^{\infty}\frac{1-x^k}{k^2}\,dx\\ &= \int_{0}^{1}\frac{1}{1-x}\sum_{k=1}^{\infty}\left ( \frac{1}{k^2}-\frac{x^k}{k^2} \right )\,dx\\ &= \int_{0}^{1}\frac{1}{1-x}\left ( \frac{\pi^2}{6}-\sum_{k=1}^{\infty}\frac{x^k}{k^2} \right )\,dx\\ &=\int_{0}^{1}\frac{\frac{\pi^2}{6}-{\rm Li_2}(x)}{1-x}\,dx \\ &=\ln(1-x)\left ( \frac{\pi^2}{6}-{\rm Li_2}(x) \right )\bigg|_0^1 + \int_{0}^{1}\frac{\ln^2 (1-x)}{x}\,dx \\ &=\int_{0}^{1}\frac{\ln^2(1-x)}{x}\,dx \\ &=2\zeta(3) \end{aligned}$$

hence the original series evaluates to $2\zeta(3)-1$. The last integral is evaluated using the Taylor expansion of $\ln(1-x)$.

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  • $\begingroup$ @user12127 Of course for the last integral you can apply parts and then use the equation ${\rm Li_2}$ satisfies and reduce it that way. It is just a matter of preference. Both lead down to the same solution. $\endgroup$ – Tolaso Feb 20 '15 at 14:57
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Asymptotic result is really easy: just notice $H_k \sim \log k$, so the resulting sum is a monotone decreasing function, hence it's asymptotically bounded by the corresponding integral: $\int_{1}^n \frac{\log x dx}{x^2}$ which is easily solved by IBP and the result is $O(\frac{\log n }{n^3})$.

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