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Note:This is the same question, but it doesn't answer my question, the answer doesn't give a closed form. In fact, the answer is not accepted. Moreover, I don't think that a 9 month old inactive question would get answers.

I have been working on various sums involving the zeta function (which come up frequently in my research), and it turned out that many of them had nice closed forms. Today, I was trying to evaluate $$\sum_{r\ge2}\frac{\zeta(r)}{r^2}$$ which turned out to be a bit harder than the others. Wolfram Alpha and Mathematica both cannot give the answer, they only give an approximate value of $0.835998$. After half an hour of work, I turned this sum to $$-\int_{0}^{\infty}\frac{t}{e^t}\psi(1-e^{-t})dt-\gamma$$ where $\gamma$ is the Euler–Mascherni constant and $\psi$ is the Digamma function. Now I don't know what to do further. Does this integral have any closed form? Any help would be appreciated. By the way, some of my ideas to evaluate the integral would be to use some integral or sum representation of the digamma function, and in this case we can interchange the sums and integrals safely.
Note: My work is too long to be stated here, so I cannot write it.
Update: This integral can be further turned to $$\lim_{n\to\infty}H_n-\gamma-\sum_{k=1}^{n}\mathrm{Li}_2\left(\frac1k\right)$$ where $\mathrm{Li}$ is the polylogarithm. I turned this to $$\lim_{n\to\infty}H_n-\gamma-\sum_{r=1}^{\infty}\frac{H_{n,r}}{r^2}$$ where $H_{n,r}$ are generalized harmonic numbers.

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  • $\begingroup$ Do you have any reason to believe this sum has a closed form? What would you do with it? $\endgroup$ – Qiaochu Yuan Dec 11 '20 at 7:34
  • $\begingroup$ @QiaochuYuan Searching for a closed form doesn't mean that we already know that there exists a closed form. Second, I have mentioned that these types of sums appear in my research, and even if they don't appear anywhere, I like evaluating sums and products for fun. $\endgroup$ – Leonhard Euler Dec 11 '20 at 7:45
  • $\begingroup$ I reversed the order of summation and got $$\sum_{n=1}^{\infty} \left [\operatorname{Li_2}{\left ( \frac1{n} \right )} - \frac1{n} \right ] $$. Unfortunately, I cannot see what to do with this, and neither does Mathematica. I have verified that this sum converges and is numerically the same as the original sum. $\endgroup$ – Ron Gordon Dec 11 '20 at 8:12
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    $\begingroup$ @RonGordon I was editing the question, to add the same, we got the same results. $\endgroup$ – Leonhard Euler Dec 11 '20 at 8:14
  • $\begingroup$ @RonGordon, the formula in your comment appears as formula (A.1) in Dr. Wolfgang Hintze's answer to the question linked to at the beginning of the OP. $\endgroup$ – Barry Cipra Dec 17 '20 at 22:57
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The simplest form i got is $$\sum_{n\ge 2}\frac{\zeta(n)}{n^2} = \int_0^1\frac{\log\Gamma(t)}{1-t}\,dt - \gamma$$ using the ordinary generating function of $\zeta(n)$.

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I've 3 different idea's, will work them out later but maybe it might be helpful.

The easiest is by using similair known generating functions e.g.: $$\sum_{m=2}^{\infty} (1+(-1)^m) \frac{\zeta(m)}{m^2}=-\int_{x=0}^{\pi} \frac{\ln\bigg(\frac{sin(x)}{x}\bigg)}{x}$$

Another one is generating functions but on the negative side of the zeta function and somehow try to extend that/find a relation. $$\int_{n=1}^{\infty} \frac{ln(n!)}{n}-\ln(n)+1 -\frac{ln(2\pi)}{2n}-\frac{\ln(n)}{2n}=-\sum_{m=1}^{\infty} \frac{\zeta(-m)}{m^2}$$

and the third one, is writing it out like you did and see if you can manipulate it there. But there are too much roads to go to know what works, just an example.

The part with 'p' goes to 0 as p goes to infinity. $$ \frac{c^n}{n^2} = \sum_{j=0}^p c^n\frac{(j)!(n-1)!}{(n+j+1)!}+\frac{(n-1)!p!}{(n+p+1)! n}$$ $$\sum_{n=1}^{\infty} \sum_{j=0}^p (j)! c^n\frac{(n-1)!}{(n+j+1)!}=$$ $$=-\ln{(c)}\ln(1-c)+\sum_{s=1}^p c^{1-s} (-1)^{s+1}\sum_{m=s}^{p}\frac{(m-1)!}{(m-s+1)!}\sum_{n=s}^m \frac{1}{n}=$$

And then adjust for the altercation ofcourse and summed over c. But there are most likely a lot of different approaches here.

And if you are lucky the sum might just show up somewhere without expecting it :)

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