How and where can I calculate constant $\sum\limits_{n=2}^{\infty}\left(\pi-n\sin(\frac{\pi}{n})\right)\approx3,132961$?

Question

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.

Answer 1

The Taylor series is simple to compute: $$ \pi-n\sin\left(\frac\pi{n}\right)=\frac{\pi^3}{6n^2}-\frac{\pi^5}{120n^4}+\frac{\pi^7}{5040n^6}-\frac{\pi^9}{362880n^8}+\dots\tag1 $$ Then we can apply the Euler-Maclaurin Sum Formula to $(1)$: $$ \begin{align} \sum_{k=2}^n\left(\pi-k\sin\left(\frac\pi{k}\right)\right) &=C-\frac{\pi^3}{6\,n}+\frac{\pi^3}{12\,n^2}-\frac{10\pi^3-\pi^5}{360\,n^3}-\frac{\pi^5}{240\,n^4}\\ &+\frac{140\pi^3+70\pi^5-\pi^7}{25200\,n^5}+\frac{\pi^7}{10080\,n^6}\\[6pt] &-\frac{10080\pi^3+3528\pi^5+252\pi^7-\pi^9}{2540160\,n^7}+O\!\left(\frac1{n^8}\right)\tag2 \end{align} $$ where $O\!\left(\frac1{n^8}\right)\approx-\frac1{25\,n^8}$

To compute $17$ places of $C$, we can plug $n=100$ into $(2)$ and get $$ C\approx3.13296163014845984\tag3 $$ To compute more places, we can use larger values of $n$, or include more terms of the asymptotic series. For example, if we use $n=2000000$ in $(2)$, we get $$ C\approx3.132961630148459846354891433606669183587072526209347\tag4 $$ Extending the asymptotic series to include terms up to $O\!\left(\frac1{n^{21}}\right)\approx\frac{16288}{n^{21}}$ and using $n=1000$, we get $$ C\approx3.1329616301484598463548914336066691835870725262093474610113\tag5 $$

Answer 2

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 $\zeta(2)e^{-3x/2}\left[1-\left(\frac{1}{24}+\frac{\pi^2}{120}\right)x^2\right]$.