MHB Summing Infinite Series with Dilogarithms

[sbhatnagar]

Hi everyone ;)

I have a challenging problem which I would like to share with you.

Prove that

\[\frac{1}{2^2}+ \frac{1}{3^2} \left(1+\frac{1}{2} \right)^2+\frac{1}{4^2} \left( 1+\frac{1}{2} +\frac{1}{3}\right)^2 + \frac{1}{5^2} \left( 1+\frac{1}{2} +\frac{1}{3}+\frac{1}{4}\right)^2 +\cdots= \frac{11\pi^4}{360}\]

 

[chisigma]

Hi everyone ;)

I have a challenging problem which I would like to share with you.

Prove that

\[\frac{1}{2^2}+ \frac{1}{3^2} \left(1+\frac{1}{2} \right)^2+\frac{1}{4^2} \left( 1+\frac{1}{2} +\frac{1}{3}\right)^2 + \frac{1}{5^2} \left( 1+\frac{1}{2} +\frac{1}{3}+\frac{1}{4}\right)^2 +\cdots= \frac{11\pi^4}{360}\]

The evaluation of...

$$S= \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{(n+1)^{2}} = \frac{11}{360}\ \pi^{4}\ (1)$$

... as well as many other 'Euler's sums' has been performed using an 'intriguing integral' by Borwein & Borwein [;)] in... http://www.math.uwo.ca/~dborwein/cv/zeta4.pdf

Kind regards

$\chi$ $\sigma$

 

[sbhatnagar]

Thank you chisigma for that nice paper. :D My solution was different from the one given in it.

I used Dilogarithms to evaluate it.