$$ \sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{(n+1)^2.2^n} \right)$$

Where $H_{n}$ is the $n^{th}$ Harmonic Number, i.e., $H_{n} = \displaystyle \sum _{k=1}^n \frac{1}{k}$

I tried to use the Integral Representation for the Harmonic number i.e.,

$$ H_{n} = \int_{0}^1 \dfrac {1-x^n}{1-x} \mathrm{d}x $$

and then interchanging the summation and integral signs, but it further complicated the problem. I also tried to use a result from my previous problem, i.e.,

$$ \sum_{n=1}^{\infty} \dfrac{1}{n^2.2^n} = \dfrac{\pi^2}{12} - \dfrac{\ln^2 2}{2} $$

but no significant progress so far.

Any help will be appreciated.


  • 1
    $\begingroup$ Have you tried just copying $H_n$'s definition and swap the two summation signs? $\endgroup$ – Alexandre Halm Mar 26 '15 at 19:57
  • $\begingroup$ @AlexHalm I don't think we can do that since in doing so, the limits will be interdependent. $\endgroup$ – user208998 Mar 26 '15 at 20:01
  • $\begingroup$ ? $ \sum_n \sum_{k \le n}$ just becomes $\sum_k \sum_{n \ge k}$. As long as everything is absolutely convergent you're good. Btw there are other ways to rearrange the sum (eg "in diagonal" $k+n=p$) $\endgroup$ – Alexandre Halm Mar 26 '15 at 20:04
  • $\begingroup$ @AlexHalm Can you please elaborate your point by writing an answer so that we can discuss more clearly? Thanks. $\endgroup$ – user208998 Mar 26 '15 at 20:09

We have$$\sum_{n\geq1}H_{n}x^{n}=-\frac{\log\left(1-x\right)}{1-x}$$ then if we integrate $$\sum_{n\geq1}\frac{H_{n}}{n+1}x^{n+1}=\frac{1}{2}\log^{2}\left(1-x\right)$$ (note we can take $C=0$ from $x=0$) then$$\sum_{n\geq1}\frac{H_{n}}{n+1}x^{n}=\frac{\log^{2}\left(1-x\right)}{2x}$$ and if we integrate again $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{2}}x^{n+1}=\int\frac{\log^{2}\left(1-x\right)}{2x}dx.$$ Now if we integrate by parts$$\int\frac{\log^{2}\left(1-x\right)}{2x}dx=\frac{\log^{2}\left(1-x\right)\log\left(x\right)}{2}+\int\frac{\log\left(1-x\right)\log\left(x\right)}{1-x}dx$$ and using the facts $\textrm{Li}'_{2}\left(1-x\right)=\frac{\log\left(x\right)}{1-x}$ and $\textrm{Li}'_{3}\left(1-x\right)=-\frac{\textrm{Li}{}_{2}\left(1-x\right)}{1-x}$ we have $$\int\frac{\log^{2}\left(1-x\right)}{2x}dx=\frac{\log^{2}\left(1-x\right)\log\left(x\right)}{2}+\log\left(x\right)\textrm{Li}{}_{2}\left(1-x\right)-\textrm{Li}{}_{3}\left(1-x\right)+C.$$ If we put $x=1$ we get $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{2}}=s_{h}\left(1,2\right)$$ which is an Euler's sum, an it can be calculated by the formula $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{m}}=\sum_{n\geq1}\frac{H_{n}}{n^{m}}-\zeta(1+m)$$ (note we have the computation $\sum_{n\geq1}\frac{H_{n}}{n^{2}}=2\zeta(3))$. So we can take $C=\zeta(3)$, then $$\sum_{n\geq1}\frac{H_{n}}{\left(n+1\right)^{2}}x^{n}=\frac{1}{x}\left(\frac{\log\left(1-x\right)\log\left(x\right)}{2}+\log\left(x\right)\textrm{Li}_{2}\left(1-x\right)-\textrm{Li}_{3}\left(1-x\right)+\zeta\left(3\right)\right).$$

  • $\begingroup$ Hi! In the last line of the obtained formula, can you tell me if we put $x=\frac{1}{2}$ (as per my question), how can we evaluate $\operatorname {Li}_{3} \left(\dfrac{1}{2}\right)$ $\endgroup$ – user208998 Mar 26 '15 at 20:41
  • $\begingroup$ @DarudeSandstorm See here en.wikipedia.org/wiki/Polylogarithm#Particular_values $\endgroup$ – Marco Cantarini Mar 26 '15 at 20:48
  • $\begingroup$ But it contains no proof or even any insights regarding it. I'd be really grateful if you could provide a simple proof too. $\endgroup$ – user208998 Mar 26 '15 at 20:52
  • $\begingroup$ @DarudeSandstorm For the dilogarithm there is the Euler reflection formula, for $\textrm{Li}{}_{3}\left(1/2\right)$ I think is possible find a proof on the internet, and maybe in this site too. $\endgroup$ – Marco Cantarini Mar 26 '15 at 22:02
  • $\begingroup$ @DarudeSandstorm You can use the identity in $(1)$, third line here mathworld.wolfram.com/Trilogarithm.html and using the facts that $\textrm{Li}{}_{3}\left(-1\right)$ it can be rewritten as Dirichlet Eta function and so as Riemann Zeta function. $\endgroup$ – Marco Cantarini Mar 26 '15 at 22:10

On the internet you can find that:

$$ \sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{n^2.2^n} \right)=\zeta(3)-\frac{\pi^2ln(2)}{12}$$

Now if we rearrange the sum : $$\begin{align}\sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{(n+1)^2.2^n} \right)&=2.\sum_{n=1}^{\infty} \left( \dfrac{H_{n}}{(n+1)^2.2^{n+1}} \right)\\ &=2\left(\zeta(3)-\frac{\pi^2ln(2)}{12}-\frac{1}{2}-\sum_{i=1}^{+\infty}\frac{1}{(n+1)^32^n}\right)\end{align}$$

and it's also known that : $$\sum_{i=1}^{+\infty}\frac{1}{n^32^n}=\frac{ln(2)^3}{6}-\frac{\pi^2ln(2)}{12}+\frac{7}{8}\zeta(3) $$

now it's your turn to rearrange all to get the result!all this sums can be found as comments in this related post.

  • 1
    $\begingroup$ Do you know a proof of the results you have used? $\endgroup$ – user208998 Mar 26 '15 at 20:22
  • $\begingroup$ Yes, but I can't write every thing here, you can just go to the link i specified in the answer and you will understand the method (it's the same for all this results), so you can solve your problem directly without passing by intermediate results! $\endgroup$ – Elaqqad Mar 26 '15 at 20:26
  • 1
    $\begingroup$ for the first one use: $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^2}=\zeta(3)+\frac{1}{2}\ln x\ln^2(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x). $$ $\endgroup$ – Elaqqad Mar 26 '15 at 20:28
  • 1
    $\begingroup$ "[B]ut I can't write every thing here, you can just go to the link [I] specified in the answer and you will understand the method." If we all had that attitude, then there would be no need for this site! So, no, "the Internet" is not a good enough answer. We are more interested in how people go about evaluating hard sums and integrals. If you have no idea how to derive these results and demonstrate you knowledge in clear language, then you may as well not bother. $\endgroup$ – Ron Gordon Mar 26 '15 at 20:30
  • $\begingroup$ @RonGordon, But in the same time we don't allow duplicates in the site (my link is internal) $\endgroup$ – Elaqqad Mar 26 '15 at 20:40

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