How and where can I calculate constant $\sum\limits_{n=2}^{\infty}\left(\pi-n\sin(\frac{\pi}{n})\right)\approx3,132961$? We can be sure, that
$$\lim\limits_{n\to\infty}n\sin(\frac{\pi}{n})=\lim\limits_{n\to\infty}n\tan(\frac{\pi}{n})=\pi$$
then we have
$$\sum\limits_{n=2}^{\infty}\left(\pi-n\sin(\frac{\pi}{n})\right)=k\approx3,132961$$
What is the type of constant is this? How and where can I calculate it more precisely?
If I made some mistakes, sorry for my English.
 A: Such constant has a not-so-terrible integral representation, which allows an accurate numerical evaluation through standard routines. For any $n\geq 2$ we have
$$ \pi-n\sin\frac{\pi}{n} = \sum_{m\geq 1}\frac{(-1)^{m+1} \pi^{2m+1}}{n^{2m}(2m+1)!}\tag{A}$$
hence by summing both sides over $n\geq 2$ we get
$$ \sum_{n\geq 2}\left(\pi-n\sin\frac{\pi}{n}\right)=\sum_{m\geq 1}\frac{(-1)^{m+1} \pi^{2m+1}\left(\zeta(2m)-1\right)}{(2m+1)!}\tag{B}$$
and by exploiting the integral representation
$$ \zeta(2m)-1 = \frac{1}{(2m-1)!}\int_{0}^{+\infty}\frac{x^{2m-1}}{e^x(e^x-1)}\,dx\tag{C}$$
we have
$$ \sum_{n\geq 2}\left(\pi-n\sin\frac{\pi}{n}\right)=\pi\int_{0}^{+\infty}\frac{\text{ber}_2\left(2\sqrt{\pi x}\right)}{xe^x(e^x-1)}\,dx \tag{D}$$
where $\text{ber}_2$ is a Kelvin function. The integrand function in the RHS of $(D)$ is positive and concentrated in a right neighbourhood of the origin, where it behaves like $\zeta(2) e^{-3x/2}$. It follows that the value of the LHS is not that far from $\frac{2\pi}{3}\zeta(2)=\frac{\pi^3}{9}=3.445\ldots$. This actually is an upper bound, and the lower bound $\frac{\pi^3(130-\pi^2)}{1215}=3.065\ldots$ can be deduced by approximating the integrand function with $\zeta(2)e^{-3x/2}\left[1-\left(\frac{1}{24}+\frac{\pi^2}{120}\right)x^2\right]$.
