# Nonlinear euler sum proof using stirling numbers of the first kind

Prove that

$$\sum_{k=1}^\infty \frac{(H_k)^2}{k^2} = \frac{17\pi^4}{360}$$

$$\textit{proof}$$

Start by the following which can be proved by induction

$$\frac{\left[n\atop 3\right]}{n!} =\frac{ (H_{n-1})^2-H^{(2)}_{n-1}}{2n}$$

And the generating function proved here

$$-\sum_{n=3}^\infty \left[n\atop 3\right] \frac{z^n}{n!} = \frac{\log^3(1-z)}{6}$$

Hence we get

$$\sum_{n=3}^\infty ( H^{(2)}_{n-1}- (H_{n-1})^2) \frac{z^n}{n}= \frac{\log^3(1-z)}{3}$$

Which implies that

$$\sum_{n=3}^\infty \frac{H^{(2)}_{n-1}- (H_{n-1})^2}{n^2}= \frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

By integration

$$\sum_{n=3}^\infty \frac{(H_{n-1})^2}{n^2}=\sum_{n=3}^\infty \frac{H^{(2)}_{n-1}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

Rewritten as

$$\sum_{n=3}^\infty \frac{(H_{n})^2}{n^2}-2\sum_{n=3}^\infty \frac{H_n}{n^3}+\sum_{n=3}^\infty \frac{1}{n^4}=\sum_{n=3}^\infty \frac{H^{(2)}_{n-1}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

Finally we have

$$\sum_{n=3}^\infty \frac{(H_{n})^2}{n^2}=2\sum_{n=3}^\infty \frac{H_n}{n^3}-2\sum_{n=3}^\infty \frac{1}{n^4}+\sum_{n=3}^\infty \frac{H^{(2)}_{n}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

All sums are shifted known terms

$$\sum_{n=1}^\infty \frac{(H_{n})^2}{n^2}=\left(H^{2}_{1}+\frac{H^{2}_{2}}{4} \right)-2\left(H_1+\frac{H_2}{8} \right)+2\left(1+\frac{1}{16}\right)-\left(H^{(2)}_{1}+\frac{H^{(2)}_{2}}{4} \right) \\+2\sum_{n=1}^\infty \frac{H_n}{n^3}-2\zeta(4)+\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$

This simplifies to

$$\sum_{n=1}^\infty \frac{(H_{n})^2}{n^2}=2\sum_{n=1}^\infty \frac{H_n}{n^3}-2\zeta(4)+\sum_{n=1}^\infty \frac{H^{(2)}_{n}}{n^2}-\frac{1}{3}\int^1_0\frac{\log^3(1-z)}{z}\,dz$$
For the Euler sums we use

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

And

$$\sum_{n= 1}^\infty \frac{H^{(k)}_n}{n^k}\, = \frac{\zeta{(2k)}+\zeta^{2}(k)}{2}$$

Then we have

$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2} = \frac{7\pi^4}{360}$$

And

$$\sum_{n=1}^\infty \frac{H_n^{(1)}}{n^3} = \frac{\pi^4}{72}$$

Finally we prove that

$$\int^1_0 \frac{\log^3(1-x)}{x}\,dx = -6\zeta(4)$$

Note that

$$\int^1_0 \frac{\log^3(1-x)}{x}\,dx = \sum_{n=0}^\infty \int^1_0 x^{n}\log^3(x)\,dx =-6\sum_{n=0}^\infty \frac{1}{(n+1)^4} = -6\zeta(4)$$

Collecting all the results we get our result

$$\sum_{n=1}^\infty \frac{(H_{n})^2}{n^2}=\frac{2\pi^4}{72}-2\zeta(4)+ \frac{7\pi^4}{360}+2\zeta(4) = \frac{17\pi^4}{360}$$

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