Sum involving zeta functions Find closed form of the following -
$$ \displaystyle \sum_{n=2}^{\infty}{\left(\frac{(n-1)\zeta(n)}{4n-1}\right)} $$
I don't know how to approach to it - Using the integral definition? I cannot use because I came to this result from there only. 
 A: One may observe that
$$
\zeta(n)=1+\frac1{2^n}+\frac1{3^n}+\cdots,\quad n>1,
$$
gives
$$
\lim_{n \to \infty}\zeta(n)=1,
$$ then 

$$
\lim_{n \to \infty}{\left(\frac{(n-1)\zeta(n)}{4n-1}\right)}=\frac14 \times 1\neq0.
$$

Your series, as written, is divergent.  
A: If you replace $\zeta(n)$ with $\zeta(n)-1$ (otherwise the series is trivially diverging, as shown by the other answers), since $\sum_{n\geq 2}\left(\zeta(n)-1\right)=1$ we just need to compute:
$$ S = \sum_{n\geq 2}\frac{\zeta(n)-1}{4n-1}. \tag{1}$$
On the other hand,
$$ \sum_{n\geq 2}\zeta(n)\, z^n = -z\left(\gamma+\psi(1-z)\right)\tag{2}$$
hence by replacing $z$ with $z^4$, then dividing by $z^2$:
$$ \sum_{n\geq 2}\zeta(n)\,z^{4n-2} = -z^2(\gamma+\psi(1-z^4))\tag{3}$$
so:
$$ \sum_{n\geq 2}\left(\zeta(n)-1\right)\,z^{4n-2} = -\gamma z^2-z^2\psi(1-z^4)-\frac{z^6}{1-z^4}\tag{4}$$
and:
$$ S = -\frac{\gamma+1}{3}-\int_{0}^{1}x^2\,\psi(2-x^4)\,dx \tag{5}$$
but honestly I do not think the last integral has a nice closed form.
A: $$\sum_{n\geq2}\frac{\zeta\left(n\right)\left(n-1\right)}{4n-1}\geq\frac{1}{4}\sum_{n\geq2}\frac{\zeta\left(n\right)}{n}\geq\frac{1}{4}\sum_{n\geq2}\frac{1}{n}.
 $$
