# A new proof a non-linear Euler sum

According to Nielsen we have the following :

If $$f(x)= \sum_{n\geq 0}a_n x^n$$

Then we have the following

$$\tag{1}\int^1_0 f(xt)\, \mathrm{Li}_2(t)\, dx=\frac{\pi^2}{6x}\int^x_0 f(t)\, dt -\frac{1}{x}\sum_{n\geq1}\frac{a_{n-1} H_{n}}{n^2}x^n$$

Now let $a_n = H_n$ then we have the following

$$f(x)=\sum_{n\geq 1}H_n x^n=-\frac{\log(1-x)}{1-x}$$

$$-\int^1_0 \frac{\log(1-xt)}{1-xt} \mathrm{Li}_2(t)\, dx=-\frac{\pi^2}{6x}\int^x_0 \frac{\log(1-t)}{1-t} dt-\sum_{n\geq1}\frac{H_{n-1} H_{n}}{n^2}x^{n-1}$$

Hence we have the following by gathering the integrals and $x\to 1$

$$\sum_{n\geq1}\frac{H_{n-1} H_{n}}{n^2}=\int^1_0\frac{\log(1-x)\left(\mathrm{Li}_2(x)-\zeta(2)\right)}{1-x} dx$$

Integrating by parts we have

$$\sum_{n\geq1}\frac{H_{n-1} H_{n}}{n^2}=-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx$$

Hence we have

$$\sum_{n\geq1}\frac{ H^2_{n}}{n^2}=\sum_{n\geq1}\frac{ H_{n}}{n^3}-\frac{1}{2}\int^1_0\frac{\log(1-x)^3}{x} dx=\frac{17 \pi^4}{360}$$

where the evaluations of the linear Euler sum and the integral could be easily found using the generating function

$$\sum_{n\geq 1}H_n x^n = -\frac{\log(1-x)}{1-x}$$

and known values for

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

• Should be $17\pi^4/360,$ not $7\pi^4/360.$ – Igor Rivin Dec 29 '13 at 20:44
• I'm not sure what the question is. – user61527 Dec 29 '13 at 20:45
• @IgorRivin, thanks I edited the answer. Actually, it is surprising that $$\sum_{n\geq 1}\frac{H_n^{(2)}}{n^2}=\frac{7\pi^4}{360}$$ – Zaid Alyafeai Dec 29 '13 at 20:51
• @T.Bongers , this not a question. Just arrived to a proof and wanted to share it with others. – Zaid Alyafeai Dec 29 '13 at 20:52
• @Ethan, thanks I will consider that. But that doesn't mean we can only post questions here or does it ? – Zaid Alyafeai Dec 29 '13 at 20:53

## 1 Answer

I mention this for mere interest, but since you are into Euler sums, have you considered learning about the contour method using the $\psi(-z)+\gamma$ kernels?.

This method can be very handy for evaluating Euler sums.

For instance, if you wanted to evaluate $\frac{H_{n}H_{n}^{(2)}}{n^{3}}$, you could try the kernel $\pi\cot(\pi z)\psi'(-z)(\gamma+\psi(-z))$.

The residue at 0 is then $\frac{13}{4}\zeta(6)+2\zeta^{2}(3)$.

In the series at the integers, n, the coefficient of the 1/(z-n) term includes $H_{n}H_{n}^{(2)}$. One can then solve for the needed sum by summing the residues. Of course, some differentiation may be required and other sums may pop up, but their known results can be used in the calculation.

Just a thought in case you have not researched this method yet.

• Thanks C. I still haven't used the contour method. It is surely useful especially for evaluating the non-linear Euler sums. – Zaid Alyafeai Dec 30 '13 at 20:41