Closed form of the sum $\sum_{r\ge2}\frac{\zeta(r)}{r^2}$ 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.
 A: 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 :)
A: 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 $\zeta(n)$.
