Consider a circle of perimeter,
\(2\pi r=10\)
a unit circle that has been scaled by
\(r=\cfrac{10}{2\pi}\).
A scaling factor of \(\cfrac{2\pi}{10}\) has to be applied to the results afterwards because of this scaling.
This way, as a integer \(n\) run along the circumference, integers overlap with a previous integer (the last digit being the same). All integer falls into \(10\) points on the circumference.
The scaling factor is necessary because on a unit circle, one unit length along the circumference extends one radian (rad) at the center. It is not true here. Scaling the unit circle to radius \(\cfrac{10}{2\pi}\), to accommodate \(10\) points on the circumference, simplifies the analysis.
Consider,
$$(\cfrac{2\pi}{10})^2\sum^{\infty}_{n=1}{\cfrac{csc^2(n.\cfrac{2\pi}{10})}{n^3}}=\cfrac{\pi^2}{25}\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}$$
where \((\cfrac{2\pi}{10})^2\) scales \((csc^2\,\,n)\) back to its length on a unit circle.
$$\sum^{\infty}_{n=1}{\cfrac{csc^2(\cfrac{n\pi}{5})}{n^3}}=\sum^\infty_{m=0}\sum^{10}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$
because when \(n=10\), \(\cfrac{n\pi}{5}=2\pi\) and the next \(10\) addends scaled by \(\cfrac{1}{n^3}\) repeats on the circle in the same position. We extract the first \(10\) terms for \(m=0\),
$$\sum^{10}_{n=1}{\cfrac{1}{n^3}}csc^2(\cfrac{n\pi}{5})$$
We are then left with eq(1),
$$\sum^\infty_{m=1}\sum^{10}_{n=1}{\cfrac{1}{(10m+n)^3}}csc^2(\cfrac{n\pi}{5})$$
Consider, when \(n=1\),
$$csc^2(\cfrac{\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+1)^3}$$
Consider, when \(n=2\),
$$csc^2(\cfrac{2\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+2)^3}=csc^2(\cfrac{2\pi}{5})\sum^\infty_{m=1}\cfrac{1}{8}\cfrac{1}{(5m+1)^3}$$
which is for all even \(n\),
$$\cfrac{1}{8}\sum^{5}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}$$
then (1) is then,
$$\sum^{5}_{i=1}csc^2(\cfrac{(2i-1)\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+(2i-1))^3} +\\ \cfrac{1}{8}\sum^{5}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}$$
And the origin summation we started with is,
$$\cfrac{\pi^2}{25}\left[\sum^{5}_{i=1}csc^2(\cfrac{(2i-1)\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(10m+(2i-1))^3} +\\ \cfrac{1}{8}\sum^{5}_{i=1}csc^2(\cfrac{2i\pi}{5})\sum^\infty_{m=1}\cfrac{1}{(5m+i)^3}+\\ \sum^{10}_{n=1}{\cfrac{1}{n^3}}csc^2(\cfrac{n\pi}{5})\right]$$
Convergent but...NOT Quite!
Addends at \(\pi\) and \(2\pi\) should cancels leaving \(8\) points on the circle, but it is not satisfying.
Note:
Flint Miller Series is $$\sum^{\infty}_{n=1}{\cfrac{csc^2(n)}{n^3}}$$
