# Compute $\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}$

How to prove, without calculating each sum separately that

$$\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=24\zeta(7)-4\zeta(2)\zeta(5)-15\zeta(3)\zeta(4)\ ?$$ where $$H_n^{(r)}=\sum_{k=1}^n\frac{1}{k^r}$$ is the generalized harmonic number and $$\zeta$$ is Riemann zeta function.

This problem is already proved by Cornel in his book Almost Impossible Integrals, Sums and Series page 299 where he used only pure series manipulations.

The question here is can we prove it in a different way?

If you are curious about the result of each sum, you can find them in the same book page 301-302.

$$\sum_{n=1}^\infty\frac{H_n^3}{n^4}=\frac{231}{16}\zeta(7)+2\zeta(2)\zeta(5)-\frac{51}{4}\zeta(3)\zeta(4)$$

$$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=-\frac{51}{16}\zeta(7)+2\zeta(2)\zeta(5)+\frac{3}{4}\zeta(3)\zeta(4)$$

Here is my proof using integration and results of some manageable Euler sums:

From here we have

$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}\right)x^n\overset{\text{reindex}}{=}\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)x^{n-1}\tag1$$

$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)x^n\tag2$$

If we multiply both sides of $$(2)$$ by $$-\frac{\ln^3x}{6x}$$ then integrate from $$x=0$$ to $$1$$

and use the fact that $$-\frac16\int_0^1 x^{n-1}\ln^3x\ dx=\frac1{n^4}$$ ,we obtain

$$\sum_{n=1}^\infty \frac{H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}}{n^4}=\frac16\int_0^1\frac{\ln^3x\ln^3(1-x)}{x(1-x)}\ dx=\frac16\mathcal I$$

Where

\begin{align} \mathcal I&=\int_0^1\frac{\ln^3x\ln^3(1-x)}{x(1-x)}\ dx\\ &=\int_0^1\frac{\ln^3x\ln^3(1-x)}{x}\ dx+\underbrace{\int_0^1\frac{\ln^3x\ln^3(1-x)}{1-x}\ dx}_{1-x\ \mapsto\ x}\\ &=2\int_0^1\frac{\ln^3x\ln^3(1-x)}{x}\ dx\overset{IBP}{=}\frac32\int_0^1\frac{\ln^4x\ln^2(1-x)}{1-x}\ dx\\ &\overset{(1)}{=}\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\int_0^1 x^{n-1}\ln^4x\ dx\\ &=\frac32\sum_{n=1}^\infty\left(H_n^2-H_n^{(2)}-\frac{2H_n}{n}+\frac2{n^2}\right)\left(\frac{24}{n^5}\right)\\ &=36\sum_{n=1}^\infty\frac{H_n^2}{n^5}-36\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}-72\sum_{n=1}^\infty\frac{H_n}{n^6}+72\zeta(7) \end{align}

Substitute the value of $$\mathcal I$$ and rearrange the terms we get

$$\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}=6\sum_{n=1}^\infty\frac{H_n^2}{n^5}-6\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}-2\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^4}-12\sum_{n=1}^\infty\frac{H_n}{n^6}+12\zeta(7)$$

We have :

$$\mathcal S_1=\sum_{n=1}^\infty\frac{H_n^2}{n^5}=6\zeta(7)-\zeta(2)\zeta(5)-\frac52\zeta(3)\zeta(4)$$

$$\mathcal S_2=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^5}=-10\zeta(7)+5\zeta(2)\zeta(5)+2\zeta(3)\zeta(4)$$

$$\mathcal S_3=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^4}=18\zeta(7)-10\zeta(2)\zeta(5)$$

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

By collecting these results, we get the closed form of the problem.

Note: $$\mathcal S_1$$ is calculated here, $$\mathcal S_2$$ and $$\mathcal S_3$$ are calculated here and $$\mathcal S_4$$ is easy to get by using Euler identity.