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Evaluate an Infinite Series

sbhatnagar

Active member
Jan 27, 2012
95
Hi everyone ;)

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

Prove that

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

chisigma

Well-known member
Feb 13, 2012
1,704
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

Active member
Jan 27, 2012
95
Thank you chisigma for that nice paper. :D My solution was different from the one given in it.

I used Dilogarithms to evaluate it.