Generalized Harmonic Number Summation ∑∞n=12−n(H(2)n)2 \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}

Prove That n=1(H(2)n)22n=1360π416π2ln22+16ln42+2Li4(12)+ζ(3)ln2

Notation : H(2)n=nr=11r2

We can solve the above problem using the generating function n=1(H(2)n)2xn, 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

starting with the integral representation of H(2)n=ζ(2)+10tnlnt1t dt we can write our sum:
S=n=1(H(2)n)22n=n=112n(ζ(2)+10xnlnx1x dx)(ζ(2)+10ynlny1y dy)=n=112n(ζ2(2)+ζ(2)10ynlny1y dy+ζ(2)10xnlnx1x dx+1010(xy)nlnxlny(1x)(1y) dx dy)
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(\color{blue}{\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}


Evaluation of the blue integral:

\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{(1-x)(2-x)}\ dx=\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{1-x}\ dx-\frac12\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{1-\frac x2}\ dx

expand \text{Li}_2(x/2) in series in the first integral and use that \sum_{n=1}^\infty H_n^{(2)}x^n=\frac{\text{Li}_2(x)}{1-x} for the second integral with replacing x by x/2

=\sum_{n=1}^\infty\frac{1}{n^22^n}\int_0^1\frac{x^n \ln x}{1-x}dx-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^n}\int_0^1 x^n\ln xdx

use (1) for the first integral

=\sum_{n=1}^\infty\frac{1}{n^22^n}\left(-\zeta(2)+H_n^{(2)}\right)-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^n}\left(-\frac1{(n+1)^2}\right)

reindex the second sum

=-\zeta(2)\text{Li}_2(1/2)+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}+\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^22^n}

write H_{n-1}^{(2)}=H_n^{(2)}-\frac1{n^2} and simplify

=-\zeta(2)\text{Li}_2(1/2)+2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-\sum_{n=1}^\infty\frac{1}{n^42^n}

=-\text{Li}_4(1/2)-\zeta(2)\text{Li}_2(1/2)+2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}

substitute

\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*}

and \text{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^2(2) , the blue closed form follows.

Attribution
Source : Link , Question Author : MathGod , Answer Author : Ali Shadhar

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