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


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


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


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:


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




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 '14 at 23:46
up vote 11 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. 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? – user 1618033 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. – user 1618033 Jun 20 '12 at 9:02

See here: (father and son)

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


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

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


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


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



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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:… – Cody Jun 30 '12 at 15:48

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