How to turn this sum into an integral?

Question

I have been trying to find the closed form of this sum to no avail. It was suggested to me to try and turn this sum into an integral and solve it like that. However, I am confused as to how to do this.

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If we have a circle of radius $r$ with an $n$-gon inscribed within this circle (i.e. with the same circumradius), we can find the difference of the areas using:

$$ A_n =\overbrace{\pi r^2}^\text{Area of circle}-\overbrace{\frac{1}{2} r^2 n \sin \left(\frac{2 \pi}{n}\right)}^\text{Area of n-gon} =r^2\left(\pi-\frac{1}{2} n \sin \left(\frac{2 \pi}{n}\right)\right) $$

I want to find the following sum (starting with $n=3$, i.e. the $n$-gon is a triangle):

$$\Lambda=\sum_{n=3}^\infty A_n = r^2\sum_{n=3}^\infty \left(\pi -\frac{1}{2} n \sin \left(\frac{2 \pi}{n}\right)\right) = r^2 \lim_{k \rightarrow \infty} \left(\pi (k-3)-\frac{1}{2} \sum_{n=3}^k n \sin \left(\frac{2 \pi}{n}\right)\right) $$

Expanding the $\sin$ using its Taylor's Series, we have

$$ \Lambda= \sum_{n=3}^\infty \pi - \frac{1}{2}n\left(\frac{2\pi}{n} - \frac{(\frac{2\pi}{n})^3}{3!} + \cdots \right) = \frac{1}{2}\sum_{n=3}^\infty \sum_{m = 1}^\infty (-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!n^{2m}} $$

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I would like some help with turning this sum into an integral to try and solve it. Any other suggestions for solving this sum are welcome as well.

My previous question on this sum can be found here: Convergence and closed form of this infinite series?

Answer 1

Here's an elaboration of my comment that yields an integral representation, if not a closed form.

I'll extend the sum to start at $n=1$ to avoid complications:

$$ \begin{align} \sum_{n=1}^\infty\sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!n^{2m}} &= \sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\sum_{n=1}^\infty n^{-2m} \\ &= \sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\zeta(2m)\tag1 \\ &= \sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\frac1{\Gamma(2m)}\int_0^\infty\frac{t^{2m-1}}{\mathrm e^t-1}\mathrm dt \\ &= \int_0^\infty\sum_{m=1}^\infty(-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!}\frac1{\Gamma(2m)}\frac{t^{2m-1}}{\mathrm e^t-1}\mathrm dt \\ &= \int_0^\infty\frac{-2\pi}{t(\mathrm e^t-1)}\sum_{m=1}^\infty\frac1{(2m-1)!(2m+1)!}\left(-(2\pi t)^2\right)^m\mathrm dt \\ &= 2\pi\int_0^\infty\frac{\operatorname{ber}_2(2\sqrt{2\pi t})}{t(\mathrm e^t-1)}\mathrm dt\;,\tag2 \end{align} $$

where the order of the summations can be changed because the double sum converges absolutely, the order of summation and integration can be changed by the dominated convergence theorem, and $\operatorname{ber}_2(t)$ is a Kelvin function. Both $(1)$ and $(2)$ can be evaluated numerically, and the results coincide; dividing by $2$ and subtracting $2\pi$ yields the approximate value $7.41721$ for your sum.