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How to prove that

$$2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}=7\zeta(7)+\frac{7}{4}\zeta(3)\zeta(4)-\frac32\zeta(2)\zeta(5)\tag{1}$$ where $H_n^{(p)}=1+\frac1{2^p}+\cdots+\frac1{n^p}$ is the $n$th generalized harmonic number of order $p$.

You can find the proof of the equality above in the book (Almost) Impossible Integrals, Sums and Series page 297 using pure series manipulations but is it possible to prove it using integration or any other way?

All approaches are appreciated.

In case you are curious about the result of each sum, you can find them also in the book $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=2\zeta(2)\zeta(5)+\frac34\zeta(3)\zeta(4)-\frac{51}{16}\zeta(7)$$

$$\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}=\frac{81}{8}\zeta(7)-\frac{11}{2}\zeta(2)\zeta(5)+\frac14\zeta(3)\zeta(4)$$

but again, our main problem here is to prove the equality in (1) in different ways.

Thanks

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In this solution, I proved

$$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)\tag{1}$$

Multiply both sides of $(1)$ by $\large \frac{\operatorname{Li}_2(x)}{x}$ then integrate from $x=0$ to $1$ and use the fact that $\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\overset{IBP}{=}\large \frac{\zeta(2)}{n}-\frac{H_n}{n^2}$ to get

$$\small{\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx-\underbrace{\int_0^1\frac{\ln(1-x)\operatorname{Li}_3(x)\operatorname{Li}_2(x)}{x}\ dx}_{IBP}-\frac12\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx}$$

$$\small{\zeta(2)\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}-\sum_{n=1}^\infty \frac{H_nH_n^{(3)}}{n^3}=\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx+\frac54\zeta(3)\zeta(4)-\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx}\tag{2}$$


By Cauchy product we have

$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)x^n\tag{3}$$

Divide both sides of $(3)$ by $x$ then integrate from $x=0$ to $1$ to get

$$\boxed{S=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)\frac1n=\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx}$$

Now multiply both sides of $(3)$ by $\large \frac{\operatorname{Li}_2(x)}{x}$ then integrate from $x=0$ to $1$ to get

$$\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx$$

$$\zeta(2)S-4\sum_{n=1}^\infty\frac{H_n^2}{n^5}-2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+6\sum_{n=1}^\infty\frac{H_n}{n^6}=\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx\tag{4}$$


By adding $(2)$ and $(4)$ and substituting the boxed value of $S=\int_0^1\frac{\operatorname{Li}^2_2(x)}{x}\ dx$ we get

$$2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}\\=\small{-\frac54\zeta(3)\zeta(4)+6\sum_{n=1}^\infty\frac{H_n}{n^6}+\zeta(2)\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty\frac{H_n^2}{n^5}+\zeta(2)\int_0^1\frac{\operatorname{Li}^2_2(x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx}$$

Now we are left with trivial integrals and lets start with the first one

\begin{align} I_1&=\int_0^1\frac{\operatorname{Li}^2_2(x)}{x}\ dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_n}{n^4} \end{align}

Similarly

\begin{align} I_2&=\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx=\sum_{n=1}^\infty\frac1{n^4}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\ &=\sum_{n=1}^\infty\frac1{n^4}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\zeta(2)\zeta(5)-\sum_{n=1}^\infty\frac{H_n}{n^6} \end{align}


Combining $I_1$ and $I_2$ gives

$$2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}\\=\frac54\zeta(3)\zeta(4)-\zeta(2)\zeta(5)-\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n^4}+7\sum_{n=1}^\infty\frac{H_n}{n^6}+\zeta(2)\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty\frac{H_n^2}{n^5}$$

We have

$$S_1=\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$

$$S_2=\sum_{n=1}^\infty\frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)$$

$$S_3=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$

$$S_4=\sum_{n=1}^\infty\frac{H_n^2}{n^5}=6\zeta(7)-\zeta(2)\zeta(5)-\frac52\zeta(3)\zeta(4)$$

By plugging the results of $S_1$, $S_2$, $S_3$ and $S_4$ we prove the equality of our problem.

Its interesting to see that the integral $\large \int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx$ got cancelled out which is really hard to crack.


Proofs: $S_1$ and $S_2$ can be found using Euler's identity, $S_4$ can be found here. As for $S_3$, we can calculate it as follows

Again,by Cauchy product we have

$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$

set $x=1$ to get

$$\zeta(2)\zeta(3)=6\sum_{n=1}^\infty\frac{H_n}{n^4}+3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-10\zeta(5)\tag{5}$$

Now lets use the well-known identity

$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty\frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$

set $p=2$ and $q=3$

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}\tag{6}$$

Plugging $(6)$ in $(5)$ and rearrange the terms we get

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)-\frac72\zeta(5)+3\sum_{n=1}^\infty\frac{H_n}{n^4}$$

Finally substitute $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ to get the closed form of $S_3.$ As a bonus, plug $S_3$ in $(6)$ to get

$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$$


BONUS:

Starting with the identity

$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^n$$

multiply both sides by $\frac{\ln^2x}{2x}$ then integrate from $x=0$ to $1$ we get

\begin{align} \sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^3}&=\frac12\int_0^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\ &=\frac12\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\underbrace{\frac12\int_0^1\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{1-x\mapsto x}\\ &=\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx=2\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1 x^n \ln^2x\ dx\\ &=4\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=4\sum_{n=1}^\infty\frac{H_n}{n^4}-4\zeta(5)=\boxed{8\zeta(5)-4\zeta(2)\zeta(3)} \end{align}

Thus $$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=8\zeta(5)-4\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}\\=\frac72\zeta(5)-\zeta(2)\zeta(3)$$

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    $\begingroup$ (+1) Artwork here. $\endgroup$ – user97357329 Aug 10 '19 at 20:43
  • $\begingroup$ @user97357329 thank you.. glad you like it. $\endgroup$ – Ali Shather Aug 10 '19 at 21:06

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