Changing Flint Miller Summation

 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 $$\sum^{\infty}_{n=1}{\cfrac{csc^2(n)}{n^3}}$$