Computing in closed form $\sum_{n=1}^{\infty}\frac{\operatorname{Ci}\left(\frac{3}{4}\zeta(2) \space n\right)}{n^2}$ What tools would you recommend me for computing the series below? 
$$\sum_{n=1}^{\infty}\frac{\operatorname{\displaystyle Ci\left(\frac{3}{4}\zeta(2) \space n\right)}}{n^2}$$
I lack the starting ideas, I need some. Thanks.
 A: $\def\cosi{{\rm Ci}}$
It is well-known, (see for instance[ 1,$\S 5.2$]), that for positive real numbers $x$ we have
$$\cosi(x)=\gamma+\ln x -\int_0^x\frac{1-\cos t}{t}\, d t.$$
It follows that, for $0<a\leq 2\pi$ and $n\geq 1$ we have
$$\cosi(a n)=\gamma+\ln a +\ln n -\int_0^a\frac{1-\cos(n t)}{t}\, d t.$$
and consequently,
$$\sum_{n=1}^\infty\frac{\cosi(a n)}{n^2}=(\gamma +\ln a)\sum_{n=1}^\infty\frac{1}{n^2}+
\sum_{n=1}^\infty\frac{\ln n}{n^2}-\int_0^a\frac{1}{t}\left(\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}\right) d t,$$
where, in the last term we used the positivity of the integrand to justify interchanging summation and integration. We conclude that
$$\sum_{n=1}^\infty\frac{\cosi(a n)}{n^2}=\frac{\pi^2}{6}(\gamma +\ln a)-
\zeta^\prime(2)-\int_0^a\frac{1}{t}\left(\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}\right) d t.\tag{1}$$
On the other hand, it is well-known, that for
$0\leq t\leq 2\pi$ that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^2}=\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4},$$
and consequently,
$$\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}=\frac{\pi t}{2}-\frac{t^2}{4}.$$
Therefore, for $0\leq a\leq 2\pi$, we obtain
$$\int_0^a\frac{1}{t}\left(\sum_{n=1}^\infty\frac{1-\cos nt}{n^2}\right)\, d t
=\int_0^a\left(\frac{\pi }{2}-\frac{t}{4}\right)\, d t=\frac{\pi a }{2}-\frac{a^2}{8}.$$
Replacing this in $(1)$ we obtain the desired formula :
$$\sum_{n=1}^\infty\frac{\cosi(a n)}{n^2}=\frac{\pi^2}{6}(\gamma +\ln a)-
\zeta^\prime(2)-\frac{\pi a }{2}+\frac{a^2}{8}, \quad\hbox{for $0<a\leq 2\pi$}$$
Remark.   This problem is enlisted as an open problem in the old Siam website here.
Remark.   Of course, I forgot to say that the case we are considering corresponds to
$a=\dfrac{3}{4}\zeta(2)=\dfrac{\pi^2}{8}$.
A: Using Maple I am obtaining $0.4848560378$ .  Do you agree?
