# Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$

Finding the closed form of:

$$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$

where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$

It appears when we try to determine the summation $\displaystyle \sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, using the generating function:

\begin{align} \sum_{n=1}^{\infty} \frac{H_n}{n^3} \, x^{n} &= - \frac{1}{2} \, \sum_{n=1}^{\infty}\frac{1}{n^2} \, \sum_{k=1}^{n} \frac{(1-x)^k}{k^2} - \frac{\zeta(2)}{2} \, \operatorname{Li}_2(x) + \frac{7 \, \zeta(4)}{8} - \frac{1}{4} \, \operatorname{Li}_2^2(1-x) + \frac{\zeta^2(2)}{4} + \operatorname{Li}_4(x) \\ & \hspace{5mm} + \frac{1}{4} \, \log^2 x \, \log^2(1-x) + \frac{1}{2}\log x \, \log (1-x) \, \operatorname{Li}_2(1-x) + \zeta(3) \, \log x - \log x \, \operatorname{Li}_2(1-x) \end{align} when we write, $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$

Combined with Cleo's closed form here, I know what the closed form should be, but how do I derive the result ?

HINT: Consider $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{2^nn^2}+\operatorname{Li}_4\left(\frac{1}{2}\right)$ and then express the remaining sum as a double integral. After some work, you get

$$\int_0^1 \frac{\displaystyle\log(x)\operatorname{Li}_2\left(\frac{x}{2}\right)}{x-2} \ dx+\operatorname{Li}_4\left(\frac{1}{2}\right)$$

and after letting $x\mapsto 2x$ combined with the integration by parts, you arrive at some integrals
pretty easy to finish. I'm confident you can finish the rest of the job to do.

$$\frac{\pi^4}{1440}-\frac{\pi^2}{3}\log^2(2)+\frac{1}{24}\log^4(2)+\frac{7}{24}\pi^2\log^2(2)+\frac{1}{4}\log(2)\zeta(3)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$

• I guess you have to develop a little bit more your answer. The 'pretty easy' is quite long. – Felix Marin Nov 28 '14 at 4:44
• @FelixMarin After I explained the details to r9m in a chat he chose my answer. I might add some details when I have time. – user 1357113 Nov 30 '14 at 22:04
• @VladimirReshetnikov You're right! Thanks. I fix that now. – user 1357113 May 23 '15 at 7:32

To find the above series, we need first to prove the following equality:

$$\displaystyle\sum_{n=1}^{\infty}H_n^{(2)}\left(\frac{\ln2}{(n+1)2^{n+1}}+\frac{1}{(n+1)^22^{n+1}}\right)=\frac1{16}\zeta(4)+\frac14 \ln^22\zeta(2)-\frac18\ln^42$$

which appeared as a problem 348 in here proposed by Cornel Ioan Valean and here is my proof:

lets start with the following integral and using the identity :$$\displaystyle\sum_{n=1}^{\infty}x^nH_n^{(2)}=\frac{\operatorname{Li_2}(x)}{1-x}$$

\begin{align*} I&=-\displaystyle \int_0^{1/2}\frac{\ln x\operatorname{Li_2}(x)}{1-x}\ dx=-\sum_{n=1}^{\infty}H_n^{(2)}\int_0^{1/2}x^n\ln x\ dx\\ &=\displaystyle\sum_{n=1}^{\infty}H_n^{(2)}\left(\frac{\ln2}{(n+1)2^{n+1}}+\frac{1}{(n+1)^22^{n+1}}\right)\tag{1} \end{align*} on the other hand : \begin{align*} I=-\int_0^{1/2}\frac{\ln x\operatorname{Li_2}(x)}{1-x}\ dx=-\int_0^{1}\frac{\ln x\operatorname{Li_2}(x)}{1-x}\ dx+\int_{1/2}^{1}\frac{\ln x\operatorname{Li_2}(x)}{1-x}\ dx \end{align*} using $$\displaystyle\sum_{n=1}^{\infty}x^nH_n^{(2)}=\sum_{n=1}^{\infty}x^{n-1}\left(H_n^{(2)}-\frac{1}{n^2}\right)=\frac{\operatorname{Li_2}(x)}{1-x}$$ for the first integral and letting $$x\mapsto1-x$$ for the second integral , we have

\begin{align*} I&=-\sum_{n=1}^{\infty}\left(H_n^{(2)}-\frac{1}{n^2}\right)\int_0^1 x^{n-1}\ln x\ dx+\overset{IBP}{\int_{0}^{1/2}\frac{\ln(1-x)\operatorname{Li_2}(1-x)}{x}\ dx}\\ &=-\sum_{n=1}^{\infty}\left(H_n^{(2)}-\frac1{n^2}\right)\left(-\frac1{n^2}\right)+\left(-\operatorname{Li_2^2}\left(\frac12\right)+\int_0^{1/2}\frac{\ln x\operatorname{Li_2}(x)}{1-x}\right)\\ 2I&=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}-\zeta(4)-\operatorname{Li_2^2}\left(\frac12\right)\tag{2} \end{align*}

Combining (1) and (2) and using $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}=\frac74\zeta(4)$$ and $$\displaystyle \operatorname{Li_2}\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^22$$ proves our equality.

now we ready to calculate the target sum : we proved \begin{align*} \frac1{16}\zeta(4)+\frac14 \ln^22\zeta(2)-\frac18\ln^42&=\sum_{n=1}^{\infty}H_n^{(2)}\left(\frac{\ln2}{(n+1)2^{n+1}}+\frac{1}{(n+1)^22^{n+1}}\right)\\ &=\sum_{n=1}^{\infty}H_{n-1}^{(2)}\left(\frac{\ln2}{n2^n}+\frac{1}{n^22^n}\right)\\ &=\ln2\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n2^n}}+\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}-\ln2\sum_{n=1}^{\infty}\frac{1}{{n^32^n}}-\sum_{n=1}^{\infty}\frac{1}{{n^42^n}} \end{align*} Rearranging the terms, we get \begin{align*} \sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\frac1{16}\zeta(4)+\frac14\ln^22\zeta(2)-\frac18\ln^42-\ln2\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n2^n}}+\ln2\operatorname{Li_3}\left(\frac12\right)+\operatorname{Li_4}\left(\frac12\right) \end{align*} Plugging $$\displaystyle\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n2^n}}=-\frac58\zeta(3)$$ and $$\displaystyle \operatorname{Li_3}\left(\frac12\right)=\frac78\zeta(3)+\frac16\ln^32-\frac12\ln2\zeta(2)$$

we get the closed form \begin{align*} \sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42 \end{align*}

• Nice answer! 🙂 If I am not mistaken manipulating the sum in question a bit leads to the similar problem AMM 11921. – r9m Apr 26 at 0:05
• @r9m thank you. I think you can relate them. Usually I prefer integral manipulation in my solutions. – Ali Shather Apr 26 at 0:17
• Btw, could you tell me which AMM problem this is? (if it's a recent one) .. I haven't followed the problem column for a while now. – r9m Apr 26 at 7:26
• @r9m sure. i edited my solution, now you can see the problem link. and sorry it was publushed on Gaceta not AMM. – Ali Shather Apr 26 at 17:46
• Many thanks... ! – r9m Apr 26 at 17:49