Contour integration problem, proving complex function is bounded I've got a complex function:
$$f(z)=\frac{\pi\cot{\pi z}}{z^2}$$
I want to integrate it around a contour $\Gamma_N$ such that the poles at $$-N,-N+1,-N+2\cdots-1,0,1,\cdots N-2, N-1, N$$ are contained in the contour, so i choose a circle with radius $N+\varepsilon$ where $N\in\Bbb{N}, \varepsilon \in (0,1)$

My goal is to show that the integral for $N\to\infty$
$$\oint_{\Gamma_N}f(z)dz=0$$
In fact, it must equal to zero because by residue theorem
$$\oint_{\Gamma_N}f(z)dz=\operatorname{Res}(f,0)+\sum_{n=-N,n\neq0}^Nn^{-2}$$
the residuum at $0$ is equal to $-\pi^2/3$ and for $N\to\infty$ this solves the basel problem. Returning to the contour integral:
$$\oint_{\Gamma_N}f(z)dz=\int_0^{2\pi}f((N+\varepsilon)e^{i\varphi})d(N+\varepsilon)e^{i\varphi}=$$
$$=\int_0^{2\pi}\frac{\pi\cot({\pi(N+\varepsilon)e^{i\varphi})}}{(N+\varepsilon)e^{i\varphi}}id\varphi$$
Now i would prove that the limit as $N\to\infty$ of the integrand goes to zero, namely:
$$\mathcal L= \lim_{N\to\infty}\frac{\cot{(\pi(N+\varepsilon))}}{(N+\varepsilon)}=\lim_{N\to\infty}\frac{\cos{(\pi(N+\varepsilon))}}{(N+\varepsilon)\sin{(\pi(N+\varepsilon)})}$$
and the argument is: $$\lim_{N\to\infty}\frac{\cos{(N+\varepsilon)}} {N+\varepsilon}=0$$ by squeeze theorem and $$\sin(\pi(N+\varepsilon))$$ is never equal to zero, because $$(N+\varepsilon)\notin\Bbb{N}$$ This would prove that integrand is zero therefore the contour integral is equal to zero. My question is, is my argument correct? Thanks for any advice.
 A: In THIS ANSWER, I showed that for $|z|+N+1/2$, the magnitude of the complex cotangent function, $|\cot(z)|$, satisfies the bound
$$|\cot(z)|=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}}\le \sqrt{1+\frac{16}{3\pi^2}}$$
Hence, we have
$$\begin{align}
\left|\,\oint_{|z|=N+1/2} \frac{\pi\cot(\pi z)}{z^2}\,dz\,\right|&\le \oint_{|z|=N+1/2}\frac{\pi |\cot(z)|}{|z|^2}\,|dz|\\\\
&\le \,2\pi (N+1/2)\left(\frac{\pi\sqrt{1+\frac{16}{3\pi^2}}}{(N+1/2)^2}\right)\\\\
&=\frac{2\pi^2\sqrt{1+\frac{16}{3\pi^2}}}{N+1/2}
\end{align}$$
which approaches $0$ as $N\to \infty$.
Finally, using the residue theorem we find that 
$$\begin{align}
\sum_{n\ne 0}\frac1{n^2}=-\text{Res}\left(\frac{\pi \cot(\pi z)}{z^2}, z=0\right)&=\lim_{z\to 0}\frac{d^2}{dz^2}\left(\pi z\cot(\pi z)\right)\\\\
&=-\frac12\lim_{z\to 0}\frac{d^2}{dz^2}\left(\frac{1-\frac12(\pi z)^2+O(\pi z)^4}{1-\frac16(\pi z)^2+O(\pi z)^4}\right)\\\\
&=-\frac12\lim_{z\to 0}\frac{d^2}{dz^2}\left(1-\frac13 (\pi z)^2+O(\pi z)^4\right)\\\\
&=\frac{\pi^2}{3}
\end{align}$$

Appendix:
In this appendix we address the question that the OP asked in a comment regarding the equality $|\cot(z)|=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}}$.  Proceeding we have
$$\begin{align}
|\cot(z)|&=\left|\frac{e^{i2z}+1}{e^{i2z}-1}\right|\\\\
&=\left|\frac{(1+\cos(2x)e^{-2y})+i\sin(2x)e^{-2y}}{(1+\cos(2x)e^{-2y})-i\sin(2x)e^{-2y}}\right|\\\\
&=\sqrt{\frac{1+2\cos(2x)e^{-2y}+\cos^2(2x)e^{-4y}+\sin^2(2x)e^{-4y}}{1-2\cos(2x)e^{-2y}+\cos^2(2x)e^{-4y}+\sin^2(2x)e^{-4y}}}\\\\
&=\sqrt{\frac{1+e^{-4y}+2\cos(2x)e^{-2y}}{1+e^{-4y}-2\cos(2x)e^{-2y}}}\\\\
&=\sqrt{\frac{\cosh(2y)+\cos(2x)}{\cosh(2y)-\cos(2x)}}
\end{align}$$
as was to be shown!
