$$\sin x = x\prod_{n=1}^\infty \left[1-\frac{x^2}{n^2\pi^2}\right]$$
If you apply $\log$ to both sides and derivate:
$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \frac{1}{1-\frac{x^2}{n^2\pi^2}}\right]$$
I have to make the expansion:
$$\frac{1}{1-\frac{x^2}{n^2\pi^2}} = 1 + {\left(\frac{x^2}{n^2\pi^2}\right)} + \left(\frac{x^2}{n^2\pi^2}\right)^2 + \left(\frac{x^2}{n^2\pi^2}\right)^3 + \cdots \tag{1}$$
But for this, I'm gonna expand the infinite sum:
$$\cot x = \frac{1}{x} - \left(\frac{2x}{1^2\pi^2} \frac{1}{1-\frac{x^2}{1^2\pi^2}} + \frac{2x}{2^2\pi^2} \frac{1}{1-\frac{x^2}{2^2\pi^2}} + \frac{2x}{3^2\pi^2} \frac{1}{1-\frac{x^2}{3^2\pi^2}}+\cdots\right)$$
So I can understand the boundaries of this expansion (cause the series $\frac{1}{1-x}=1 + x + x^2 + x^3 + \cdots$ as long as $|x|<1$.
So, am I right in saying that I can do the expansion $(1)$ as long as $|x|<\pi$? Because for $x=\pi$ we have: $$\frac{1}{1-\frac{\pi^2}{1^2\pi^2}} = \frac{1}{1-1}$$ And if $|x|<\pi$ the other terms like:
$$\frac{1}{1-\frac{\pi^2}{2^2\pi^2}}$$ Can be expanded by $(1)$ too.
So:
$$\cot x = \frac{1}{x} - \sum_{n=1}^\infty \left[\frac{2x}{n^2\pi^2} \left(1 + {\left(\frac{x^2}{n^2\pi^2}\right)} + \left(\frac{x^2}{n^2\pi^2}\right)^2 + \left(\frac{x^2}{n^2\pi^2}\right)^3 + \cdots \right)\right]\tag{$|x|<\pi$}$$ $$x \cot x = 1 - 2\sum_{n=1}^\infty \left[\frac{x^2}{n^2\pi^2} + \frac{x^4}{n^4\pi^4} + \frac{x^6}{n^6\pi^6} + \frac{x^8}{n^8\pi^8} + \cdots\right]$$
And finally the famous: $$x \cot x = 1 - 2\sum_{n=1}^{\infty} \left[\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right]\tag{$|x|<\pi$}$$
Am I right with the boundaries for $x$?