Starting with $$\sin{nx}=2^{n-1}\prod^{n-1}_{k=0}\sin{(x+\frac{\pi k}{n})}$$ take logarithm, then derivative, then substitute $x$ with $\pi x/n$, to get $$\pi \cot{\pi x} = \sum^{n-1}_{k=0}\frac{\pi}{n}\cot{\big(\frac{\pi}{n}(x+k)\big)}$$ Thenceforth restrict $x\in (0,1)$.
Substitute $n$ with $2n$, then fold the summation in half, $$\pi \cot{\pi x} = \sum^{n-1}_{k=0}\frac{\pi}{2n}\Big(\cot{\big(\frac{\pi}{2n}(x+k)\big)}+\cot{\big(\frac{\pi}{2n}(x+(2n-1-k))\big)}\Big)$$ $$= \sum^{n-1}_{k=0}\frac{\pi}{2n}\Big(\cot{\big(\frac{\pi}{2n}(x+k)\big)}-\cot{\big(\frac{\pi}{2n}(k+1-x)\big)}\Big)$$
Then we estimate $\cot{\big(\frac{\pi}{2n}(x+k)\big)}-\cot{\big(\frac{\pi}{2n}(k+1-x)\big)}$ at two extremes.
When $k\sim n$, expand it as a power in $x, (1-x)$ to $$=\frac{1}{\sin^2{(\frac{\pi k}{2n})}}\Big(\frac{\pi}{2n}(1-2x)+O(1/n^2)\Big)$$
Thus $\forall 0<\theta<1$, $$\sum^{n-1}_{k=\theta n}\frac{\pi}{2n}\Big(\cot{\big(\frac{\pi}{2n}(x+k)\big)}-\cot{\big(\frac{\pi}{2n}(k+1-x)\big)}\Big)$$ $$\sim\sum^{n-1}_{k=\theta n}(\frac{\pi}{2n})^2 \frac{1}{\sin^2{(\frac{\pi k}{2n})}} = O(1/n) \rightarrow 0$$ So as $n\rightarrow \infty$, the summation is concentrted into the $0\leq k \leq \theta n$ part.
When $0\leq k \leq \theta n$, if we expand by $\cot{t}=1/t-t/3+O(t^3)$, and discard all but the first term, we get $$\sum^{\theta n}_{k=0}\frac{\pi}{2n}\Big(\cot{\big(\frac{\pi}{2n}(x+k)\big)}-\cot{\big(\frac{\pi}{2n}(k+1-x)\big)}\Big)\approx \sum^{\theta n}_{k=0} \frac{1}{x+k}+\frac{1}{x-(k+1)}=\sum^{\theta n}_{k=-\theta n} \frac{1}{x+k}$$
If this is correct, then taking $n\rightarrow \infty$ would yield $\pi\cot{\pi x}=\sum_{-\infty}^{\infty}\frac{1}{x+k}$.
But I know that it has subtlety. In particular, I think the error in discarding all higher powered terms depends both on $n$ and on $\theta$, and that merely letting $n\rightarrow \infty$ would not make the error disappear. That is, I think $$\sum^{\theta n}_{k=0}\frac{\pi}{2n}\Big(\cot{\big(\frac{\pi}{2n}(x+k)\big)}-\cot{\big(\frac{\pi}{2n}(k+1-x)\big)}\Big)=\sum^{\theta n}_{k=-\theta n} \frac{1}{x+k}+E(n, \theta)$$ where $\lim_{n\to \infty}E(n, \theta)$ might not be $0$, but $\lim_{\theta\to 0}\lim_{n\to \infty}E(n, \theta)=0$
That's where matter stands now. All I need is to get a good estimation for the $E(n, \theta)$. Can you help me find a bound to $E$?