# Prove the Harmonic sum

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the following

$$\displaystyle \sum_{k\geq 1} \frac{H^2_k}{k^2}=\frac{17}{4}\zeta(4)=\frac{17\pi^4}{360}$$

$$\displaystyle \mbox{where }\,\,H^2_k =\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{k}\right)^2$$​

#### chisigma

##### Well-known member
Prove the following

$$\displaystyle \sum_{k\geq 1} \frac{H^2_k}{k^2}=\frac{17}{4}\zeta(4)=\frac{17\pi^4}{360}$$

$$\displaystyle \mbox{where }\,\,H^2_k =\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{k}\right)^2$$​
The prove was supplied in the year 1995 by D. Borwein and J. M. Borwein in the article On an intriguing integral and some series related to $\zeta (4)$...

http://www.ams.org/journals/proc/1995-123-04/S0002-9939-1995-1231029-X/S0002-9939-1995-1231029-X.pdf

Kind regards

$\chi$ $\sigma$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Actually , I haven't even tried it . I was reading the paper and thought it is a good exercise. I will look for a solution of my own .

#### topsquark

##### Well-known member
MHB Math Helper
Actually , I haven't even tried it . I was reading the paper and thought it is a good exercise. I will look for a solution of my own .
Ouch! I thought that paper was pretty brutal. I could follow the work on the integral but got lost on the Corollary work. I was kind of hoping you had a newer (and simpler) view on the matter.

-Dan

Staff member

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Zaid,

When a problem is posted here in the Challenge Questions and Puzzles sub-forum, it is expected that you already have a solution ready to post after giving our members a fair amount of time to solve it, as per http://www.mathhelpboards.com/f28/guidelines-posting-answering-challenging-problem-puzzle-3875/.
I know the solution , it is in the paper . But it is very long and requires lots of things to work out . I was hoping for someone to post a newer method .

If that doesn't suit here , you can move the topic to an appropriate section .