Finite sum $\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2(\frac{\pi x}{n})$

Solution 1:

An idea:

Note the identity

$$\sin n \theta=U_{n-1}\left( \cos \theta \right) \sin \theta ,$$

where $U_i$ is the $i^{th}$ Chebyshev orthogonal polynomial of the second kind. This gives

$\sin^2 \pi x=\sin^2 \left(n \frac{\pi x}{n}\right)=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n},$

and then

$\sin^2 \pi x \csc^2 \frac{\pi x}{n}=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right) \sin^2 \frac{\pi x}{n} \csc^2 \frac{\pi x}{n}=U_{n-1}^2\left( \cos \frac{\pi x}{n} \right),$

so your sum becomes

$\sum \limits_{n=2}^N \frac{1}{n^2} U_{n-1}^2\left( \cos \frac{\pi x}{n} \right).$

There may be an identity on the Chebyshev polynomials to further simplify this sum.