Compute $\sum_{n=1}^\infty\frac{H_n^4}{n^2}$ and $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}$

Question

How to prove that

$$\sum_{n=1}^\infty\frac{H_n^4}{n^2}=\frac{979}{24}\zeta(6)+3\zeta^2(3)\ ?$$ $$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)\ ?$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

We can find these two sums nicely evaluated in the book Almost Impossible Integrals, Sums and series page 429 using pure series manipulations and I managed to find their closed forms using integration and some harmonic-number identities ( Solution will be posted in the answer section ), but can we tackle them by other means like contour integration or the $\arcsin^4x$ identity just like what @nospoon did in his solution here, where he mentioned the result of the second sum in $(4)$ but not the first one, or by any other way?

Thanks .

Answer 1

To compute these two sums, we are going to establish two relations and solve them by elimination.

First Relation:

From here we have $$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$

Multiply both sides by $\frac{H_n}{n}$ then sum both sides from $n=1$ to $\infty$ to get

\begin{align} R_1&=\sum_{n=1}^\infty\frac{H_n^4}{n^2}+3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=-\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty\frac{H_n}{n}x^n\ dx\\ &=-\int_0^1\frac{\ln^3(1-x)}{x}\left(\frac12\ln^2(1-x)+\operatorname{Li}_2(x)\right)\ dx\\ &=-\frac12\int_0^1\frac{\ln^5(1-x)}{x}\ dx-\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_2(x)}{x}\ dx\\ &=60\zeta(6)-I \end{align}

Then $$\boxed{R_1=\sum_{n=1}^\infty\frac{H_n^4}{n^2}+3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=60\zeta(6)-I}$$.

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Second Relation:

From here, we have

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Divide both sides by $x$ then integrate from $x=0$ to $x=y$, we get

$$-\int_0^y\frac{\ln^3(1-x)}{x(1-x)}\ dx=\sum_{n=1}^\infty \frac{y^n}{n}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$

Now multiply both sides by $-\frac{\ln(1-y)}{y}$ then integrate from $y=0$ to $y=1$ and use the fact that $-\int_0^1 y^{n-1}\ln(1-y)\ dy=\frac{H_n}{n}$, we get

\begin{align} R_2&=\sum_{n=1}^\infty\frac{H_n^4}{n^2}-3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=\int_0^1\int_0^y\frac{\ln^3(1-x)\ln(1-y)}{xy(1-x)}\ dx\ dy\\ &=\int_0^1\frac{\ln^3(1-x)}{x(1-x)}\left(\int_x^1\frac{\ln(1-y)}{y}\ dy\right)\ dx\\ &=\int_0^1\frac{\ln^3(1-x)}{x(1-x)}\left(\operatorname{Li}_2(x)-\zeta(2)\right) dx\\ &=\int_0^1\frac{\ln^3(1-x)}{x}\left(\operatorname{Li}_2(x)-\zeta(2)\right) dx+\underbrace{\int_0^1\frac{\ln^3(1-x)}{1-x}\left(\operatorname{Li}_2(x)-\zeta(2)\right) dx}_{IBP}\\ &=I-\zeta(2)\int_0^1\frac{\ln^3(1-x)}{x}\ dx-\frac14\int_0^1\frac{\ln^5(1-x)}{x}\ dx\\ &=I+6\zeta(2)\zeta(4)+30\zeta(6)\\ &=I+\frac{81}{2}\zeta(6) \end{align}

Then $$\boxed{R_2=\sum_{n=1}^\infty\frac{H_n^4}{n^2}-3\sum_{n=1}^\infty\frac{H_n^2 H_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}=I+\frac{81}{2}\zeta(6)}$$.

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Therefore

\begin{align} \sum_{n=1}^\infty\frac{H_n^4}{n^2}&=\frac{R_1+R_2}{2}=\frac{201}{4}\zeta(6)-2\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}\\ &=\frac{201}{4}\zeta(6)-2\left(\frac{227}{48}\zeta(6)-\frac32\zeta^2(3)\right)\\ &=\frac{797}{24}\zeta(3)+3\zeta^2(3) \end{align}

where the value of $\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^2}$ is calculated here.

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and

\begin{align} \sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}&=\frac{R_1-R_2}{6}=\frac{13}{4}\zeta(6)-\frac13I \end{align}

and the remaining integral can be approached as follows:

\begin{align} I&=\int_0^1\frac{\ln^3(1-x)\operatorname{Li}_2(x)}{x}\ dx\overset{1-x\ \mapsto\ x}{=}\int_0^1\frac{\ln^3x\operatorname{Li}_2(1-x)}{1-x}\ dx\\ &=\int_0^1\frac{\ln^3x}{1-x}(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x))\ dx\\ &=\zeta(2)\int_0^1\frac{\ln^3x}{1-x}\ dx-\int_0^1\frac{\ln^4x\ln(1-x)}{1-x}\ dx-\int_0^1\frac{\ln^3x\operatorname{Li}_2(x)}{1-x}\ dx\\ &=-6\zeta(2)\zeta(4)+\sum_{n=1}^\infty H_n\int_0^1x^n\ln^4x\ dx-\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n \ln^3x \ dx\\ &=-\frac{21}2\zeta(6)+24\sum_{n=1}^\infty\frac{H_n}{(n+1)^5}+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^4}\\ &=-\frac{21}2\zeta(6)+24\sum_{n=1}^\infty\frac{H_n}{n^5}-24\zeta(6)+6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}-6\zeta(6)\\ \end{align}

Substituting $\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^4}=\zeta^2(3)-\frac13\zeta(6)$ ( proved here), we get

$$I=-\frac12\zeta(6)-6\zeta^2(3)$$

Consequently $$\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^2}=\frac{41}{12}\zeta(6)+2\zeta^2(3)$$