Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$

Question

Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$

Notation : $ \displaystyle H_{n}^{(2)} = \sum_{r=1}^{n} \dfrac{1}{r^2}$

We can solve the above problem using the generating function $\displaystyle \sum_{n=1}^{\infty} (H_{n}^{(2)})^2 x^n $, but it gets rather tedious especially taking into account the indefinite polylogarithm integrals involved. Can we solve it using other methods like Euler Series Transform or properties of summation?

Answer 1

starting with the integral representation of $\displaystyle H_n^{(2)}=-\zeta(2)-\int_0^1\frac{t^n\ln t}{1-t}\ dt$, we can write our sum: \begin{align} S&=\sum_{n=1}^\infty\frac{\left(H_n^{(2)}\right)^2}{2^n}=\sum_{n=1}^\infty\frac1{2^n}\left(-\zeta(2)-\int_0^1\frac{x^n\ln x}{1-x}\ dx\right)\left(-\zeta(2)-\int_0^1\frac{y^n\ln y}{1-y}\ dy\right)\\ &=\sum_{n=1}^\infty\frac1{2^n}\left(\zeta^2(2)+\zeta(2)\int_0^1\frac{y^n\ln y}{1-y}\ dy+\zeta(2)\int_0^1\frac{x^n\ln x}{1-x}\ dx+\int_0^1\int_0^1\frac{(xy)^n\ln x\ln y}{(1-x)(1-y)}\ dx\ dy\right) \end{align} note that the second and the third term have the same value and using the geometric series, we have

\begin{align} S&=\zeta^2(2)+2\zeta(2)\int_0^1\frac{\ln x}{1-x}\sum_{n=1}^\infty\left(\frac{x}{2}\right)^n\ dx+\int_0^1\int_0^1\frac{\ln x\ln y}{(1-x)(1-y)}\sum_{n=1}^\infty\left(\frac{xy}{2}\right)^n\ dx\ dy\\ &=\zeta^2(2)+2\zeta(2)\int_0^1\frac{x\ln x}{(1-x)(2-x)}\ dx+\int_0^1\int_0^1\frac{xy\ln x\ln y}{(1-x)(1-y)(2-xy)}\ dx\ dy\\ &=\zeta^2(2)+2\zeta(2)(-\ln^22)+\int_0^1\frac{\ln x}{1-x}\left(\int_0^1\frac{xy\ln y}{(1-y)(2-xy)}\ dy\right)\ dx\\ &=\zeta^2(2)-2\zeta(2)\ln^22+\int_0^1\frac{\ln x}{(1-x)(2-x)}\left(\int_0^1\frac{x\ln y}{1-y}\ dy-\int_0^1\frac{2x\ln y}{2-xy}\ dy\right)\ dx\\ &=\zeta^2(2)-2\zeta(2)\ln^22+\int_0^1\frac{\ln x}{(1-x)(2-x)}\left(-\zeta(2)x+2\operatorname{Li_2}\left(\frac{x}{2}\right)\right)\ dx\\ &=\zeta^2(2)-2\zeta(2)\ln^22+(-\ln^22)(-\zeta(2))+2\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{(1-x)(2-x)}\ dx\\ &=\zeta^2(2)-\zeta(2)\ln^22+2\color{blue}{\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{(1-x)(2-x)}\ dx}\\ &=\frac52\zeta(4)-\zeta(2)\ln^22+2\left(\operatorname{Li_4}\left(\frac{1}{2}\right)-\frac98\zeta(4)+\frac12\ln2\zeta(3)+\frac1{12}\ln^42\right)\\ &=2\operatorname{Li_4}\left(\frac{1}{2}\right)+\frac14\zeta(4)+\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42 \end{align}

I can provide the proof of the blue integral if someone is interested

Answer 2

second approach suggested by Cornel Ioan Valean using summation by parts and lets start with the following sum:

with ${N \in \mathbb{N}_{\ \geq\ 1}}$ \begin{align} \sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^N\frac{\left(H_n^{(2)}\right)^2}{2^n}-2\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}+\sum_{n=1}^N\frac1{n^42^n}\tag{1} \end{align} on the other hand: \begin{align} \sum_{n=1}^N\frac{\left(H_{n-1}^{(2)}\right)^2}{2^n}=\sum_{n=1}^{N-1}\frac{\left(H_{n}^{(2)}\right)^2}{2^{n+1}}=\sum_{n=1}^{N}\frac{\left(H_{n}^{(2)}\right)^2}{2^{n+1}}-\frac{\left(H_{N}^{(2)}\right)^2}{2^{N+1}}\tag{2} \end{align} from $(1)$ and $(2)$ we reach $$\sum_{n=1}^N\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=4\sum_{n=1}^N\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^N\frac{1}{n^42^n}-2\frac{\left(H_{N}^{(2)}\right)^2}{2^{N+1}}$$ letting $N$ approach $\infty$ we get $$\sum_{n=1}^\infty\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=4\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-2\sum_{n=1}^\infty\frac{1}{n^42^n}-0$$ I was able here to prove $$\begin{align*} \sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42 \end{align*}$$ which follows $$\sum_{n=1}^\infty\frac{\left(H_{n}^{(2)}\right)^2}{2^n}=2\operatorname{Li_4}\left(\frac{1}{2}\right)+\frac14\zeta(4)+\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42 $$