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I ran across a fun looking series and am wondering how to tackle it.

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=\frac{{\pi}^{4}}{72}.$$

One idea I had was to use the digamma and the fact that

$$\sum_{k=1}^{n}\frac{1}{k}=\int_{0}^{1}\frac{1-t^{n}}{1-t}dt=\psi(n+1)+\gamma.$$

Along with the identity $\psi(n+1)=\psi(n)+\frac{1}{n}$, I managed to get it into the form

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=\gamma\zeta(3)+\zeta(4)+\sum_{n=1}^{\infty}\frac{\psi(n)}{n^{3}}.$$

This would mean that $$\sum_{n=1}^{\infty}\frac{\psi(n)}{n^{3}}=\frac{{\pi}^{4}}{360}-\gamma\zeta(3).$$ Which, according to Maple, it does. But, how to show it?. If possible.

I also started with $\frac{-\ln(1-x)}{x(1-x)}=\sum_{n=1}^{\infty}H_{n}x^{n-1}$.

Then divided by x and differentiated several times. This lead to some interesting, but albeit, tough integrals involving the dilog:

$$-\int\frac{\ln(1-x)}{x(1-x)}dx=Li_{2}(x)+\frac{\ln^{2}(1-x)}{2}=\sum_{n=1}^{\infty}\frac{H_{n}x^{n}}{n}.$$

Doing this again and again lead to some integrals that appeared to be going in the right direction.

$$\int_{0}^{1}\frac{Li_{3}(x)}{x}dx=\frac{{\pi}^{4}}{90}$$

$$-\int_{0}^{1}\frac{\ln^{2}(1-x)\ln(x)}{2x}dx=\frac{{\pi}^{4}}{360}$$

$$-\int_{0}^{1}\frac{\ln(1-x)Li_{2}(1-x)}{x}dx=\frac{{\pi}^{4}}{72}$$

But, what would be a good approach for this one? I would like to find out how to evaluate

$$\sum_{n=1}^{\infty}\frac{\psi(n)}{n^{3}}=\frac{{\pi}^{4}}{360}-\gamma\zeta(3)$$ if possible, but any methods would be appreciated and nice.

Thanks a bunch.

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There is a proof of the first result from above by harmonic sum methods and Mellin transforms at the following MSE link. –  Marko Riedel Mar 23 at 23:46

3 Answers 3

up vote 10 down vote accepted

$$\sum_{n=1}^{+\infty} \frac{H_{n}}{n^{3}} = \sum_{n=1}^{+\infty} \frac{1}{n^{3}} \sum_{m=1}^{+\infty} \left( \frac{1}{m} - \frac{1}{m+n}\right) = \sum_{n=1}^{+\infty} \frac{1}{n^{3}} \sum_{m=1}^{+\infty} \frac{n}{m(m+n)} = \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{m}{m^2 n^2 (m+n)} = \frac{1}{2} \left(\sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{m}{m^2 n^2(m+n)} + \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{n}{m^2 n^2(m+n)} \right) = \frac{1}{2} \sum_{n=1}^{+\infty} \sum_{m=1}^{+\infty} \frac{1}{m^2 n^2} = \frac{1}{2} \zeta(2)^2 = \frac{1}{2} \left(\frac{\pi^{2}}{6}\right)^2 = \frac{\pi^{4}}{72} = \frac{5}{4} \zeta(4) $$

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1  
+1. Nice and elegant! –  user17762 Jun 19 '12 at 18:47
    
@qoqosz:nice. You have my vote! Is it possible to solve it without doubling the sum? –  Chris's sis Jun 19 '12 at 21:22
    
Yes indeed, a very nice and elegant method. Thanks. –  Cody Jun 19 '12 at 21:34
    
@Chris you mean double summation $\sum_n \sum_m$? I don't think so, because when treating this sum elementary $H_n$ is sum itself. –  qoqosz Jun 20 '12 at 8:50
    
@qoqosz: yeah. I noticed that it's hard to go other way. –  Chris's sis Jun 20 '12 at 9:02

I appreciate all of the input.

I thought I would come back and post something I managed to come up with.

This is kind of based on the methods in my first post using the dilog.

I started by using the identity $-n\int_{0}^{1}(1-x)^{n-1}\ln(x)dx=-\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k}}{k}=H_{n}$.

Then, $\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=-\sum_{n=1}^{\infty}\frac{1}{n^{2}}\int_{0}^{1}(1-x)^{n-1}\ln(x)dx$

$=-\int_{0}^{1}\sum_{n=1}^{\infty}\frac{(1-x)^{n-1}\ln(x)}{n^{2}}dx$

Using the definition of the dilog, $Li_{2}(1-x)=\sum_{n=1}^{\infty}\frac{(1-x)^{n}}{n^{2}}$, I got:

$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{3}}=-\int_{0}^{1}\frac{Li_{2}(1-x)\ln(x)}{1-x}dx$

$=\frac{1}{2}(Li_{2}(1-x))^{2} |_{0}^{1}$

$=\frac{1}{2}(Li_{2}(1))^{2}=\frac{1}{2}\left(\frac{{\pi}^{2}}{6}\right)^{2}$

$=\frac{{\pi}^{4}}{72}$.

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Sorry Nick. I realize this is rather belated, but in fairness to Nick Strehlke, sometime back he provided a nice solution to the question at hand in this post. Thanks Nick. It's cool how you related these topics: math.stackexchange.com/questions/119253/… –  Cody Jun 30 '12 at 15:48

See here: (father and son)

On An Intriguing Integral and Some Series Related to $\zeta(4)$ - David Borwein and Jonathan M. Borwein

Enjoy

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Thanks for the link. I searched around, but did not locate that one. –  Cody Jun 19 '12 at 21:35

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