complex analysis - A difficult functional series (still unsolved)

The problem is to prove if the following functional series converges or not and in the affirmative case to find its sum

$$\qquad\qquad\ \sin^2(\pi x) \sum_{k=1}^\infty\frac{1}{k^2\sin^2(\frac{\pi x}{k})}=\sum_{k=1}^\infty \frac{\sin^2(\pi x)}{k^2\sin^2(\frac{\pi x}{k})}\qquad\ \text{with}\quad x,k\in\Bbb N-\{0\}$$

The denominator of the series, upon which the summation variable acts, can be rewritten in many ways, as follows, none of which have seemed to be useful to solve the problem:

$$\frac{1}{k^2\sin^2(\frac{\pi x}{k})}=\frac{1}{(\pi x)^2\operatorname{sinc}^2(\frac{\pi x}{k})}=\frac{\csc^2(\frac{\pi x}{k})}{k^2}=\frac{1+\cot^2(\frac{\pi x}{k})}{k^2}=\frac{2}{k^2(1-\cos(\frac{2\pi x}{k}))}=\frac{1}{2k^2(1+\cos(\frac{\pi x}{k}))}+\frac{1}{2k^2(1-\cos(\frac{\pi x}{k}))}$$

where
$$ \csc(x)=\frac{1}{\sin(x)}\quad\text{and}\quad \operatorname{sinc}(x)=\frac{\sin(x)}{x}$$

Also

$$\frac{1}{k^2\sin^2(\frac{\pi x}{k})}=\frac{1}{\pi^2}\sum_{m=-\infty}^\infty \frac{1}{(x-mk)^2}=\frac{1}{\pi^2}\sum_{m=-\infty}^\infty \frac{1}{(x+mk)^2}$$

since

$$\frac{1}{\sin^2(x)}=\csc^2(x)=\sum_{m=-\infty}^\infty\frac{1}{(x-m\pi)^2}=\sum_{m=-\infty}^\infty\frac{1}{(x+m\pi)^2}$$

Maybe it should be worth noting that the series considered resembles in some way the so called Flint Hills Series (that, differently from the one considered, is a numerical series: http://mathworld.wolfram.com/FlintHillsSeries.html) for which, unto this day, it is not known whether it converges or not.

Note:

It is obvious that the result is always $0$, and hence trivial, if the function is evaluated before performing the sum, but this is not what is intended in this case.

In fact, for example, the value resulting for $x=10$ after that the summation of

$$\sum_{k=1}^N \frac{\sin^2(\pi x)}{k^2\sin^2(\frac{\pi x}{k})}\qquad\ \text{with}\quad x,k,N\in\Bbb N-\{0\}$$

has been performed up to $N\geqslant x=10$ is, and will always be (meaning that it stands for $N \to \infty$), $4$.

Summarizing, it can be easily verified (for convenience with a software like Maple, for example) that

$$\forall\alpha\in\Bbb N-\{0\}\qquad\lim_{x\to\alpha}\Biggl(\sum_{k=1}^{N\geqslant \alpha} \frac{\sin^2(\pi x)}{k^2\sin^2(\frac{\pi x}{k})}\Biggr)=\beta\neq0\qquad\ \text{with}\quad \beta\in\Bbb N-\{0\}$$
where, clearly, this is one of those cases in which the operation of limit and summation cannot be interchanged.
Another example, but that can be carried out by hand:

$$\lim_{x\to 2}\Biggl(\sum_{k=1}^2\frac{\sin^2(\pi x)}{k^2\sin^2(\frac{\pi x}{k})}\Biggr)=\lim_{x\to 2}\Biggl(1+\frac{\sin^2(\pi x)}{2(1-\cos(\pi x))}\Biggr)=2\neq0$$

So, from the calculations of the partial sums up to an arbitrary extent is evident that, numerically, the summation converge (pointwise?); the problem is to see if it is possible to find, first, an analytical closed form for the summation.