<s>[INST] Prove convergence of trignometric sum from $-\infty$ to $\infty$ 
Prove convergence of $$\pi^2\sum_{n=-\infty}^{\infty}\left ( \frac{1}{\cos^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}-\frac{1}{\sin^2\left (\pi\left (n-\frac{1}{2}\right )\tau\right )}\right )$$ and $$\pi^2\sum_{n=-\infty}^{\infty}\left ( \frac{1}{\cos^2\left (\pi n\tau \right )}-\frac{1}{\sin^2\left (\pi\left (n-\frac{1}{2}\right )\tau\right )}\right )$$
  where $\tau$ is a constant.

The proof I have which I am trying to follow goes like this (Ahlfors text): 

The series are strongly convergent for both $n\rightarrow +\infty$ and $n\rightarrow -\infty$ for $|\cos n\pi\tau|$ and $|\sin n\pi\tau|$ are comparable to $e^{|n|\pi\text{Im}(\tau)}$ and hence the convergence is uniform for $\text{Im}(\tau)\geq \delta>0$. We can now take the limits termwise and we find that the first sum converges to zero whilst the second conerges to $\pi^2$ (from the $n=0$ term).

I don't understand most of the proof; I can see how changing to $e^{|n|\pi\text{Im}(\tau)}$ satisfies convergence for $\text{Im}(\tau)>0$. However I don't see how that relates to $\cos$ and $\sin$ and why taking limits termiwise gives the result it does (I understand we are allowed to as we have uniform convergence).
 [/INST] (1) How $e^{|n|\pi\text{Im}(\tau)}$ relates to $\cos$ and $\sin$ :
 For simplicity we first argue about $$\sum_{n=1}^\infty \frac{1}{\cos^2 \left (\pi\left (n-\frac{1}{2}\right )\tau \right )}.$$
Since $$\left|e^{i\pi\left(n-\frac{1}{2}\right)\tau}\right|=
e^{-\pi\left(n-\frac{1}{2}\right)\sigma}
\quad\text{and}\quad 
\left|e^{-i\pi\left(n-\frac{1}{2}\right)\tau}\right|=
e^{\pi\left(n-\frac{1}{2}\right)\sigma}
,$$ where $\sigma=\text{Im}(\tau)>0$, we have 
\begin{align}
2\left|\cos \left (\pi\left (n-\frac{1}{2}\right )\tau \right )\right|&=\left|e^{i\pi\left(n-\frac{1}{2}\right)\tau}+ e^{-i\pi\left(n-\frac{1}{2}\right)\tau}      \right|\\
&\ge e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}.
\end{align}
Furthermore we have
\begin{align}
2\left|\cos \left (\pi\left (n-\frac{1}{2}\right )\tau \right )\right|&\ge e^{\pi\left(n-\frac{1}{2}\right)\sigma} \left(1-  e^{-\pi(2n-1)\sigma}    \right)\\
&\ge \frac{1}{2} e^{\pi\left(n-\frac{1}{2}\right)\sigma}\ge \frac{1}{2} e^{\pi(n-1)\sigma}
\end{align}
 for $n\ge n_0,$ where $n_0$ is an integer such that $e^{-(2n_0-1)\pi\sigma}\le \frac{1}{2}$. 
Therefore $$
\sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|
\le \sum_{n=1}^{n_0}  \frac{4}{ \left(e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}\right)^2 }  +\sum_{n=n_0+1}^\infty 4^2\cdot \,e^{-2(n-1)\pi\sigma}<\infty,
$$
which ensures the absolute convergence of $\sum_{n=1}^\infty 1/\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )$.  
(2) Taking limits termwise or Uniform convergence:
If $\sigma\ge \delta >0$, we can take $n_0$ (depending only on $\delta $) for arbitrary $\varepsilon >0$  so that $$
\sum_{n=n_0+1}^\infty 4^2\cdot \,e^{-2(n-1)\pi\sigma}\le \sum_{n=n_0+1}^\infty 4^2\cdot \,e^{-2(n-1)\pi\delta }<\varepsilon.
$$
Then we have $$
\sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|
\le \sum_{n=1}^{n_0}  \frac{4}{ \left(e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}\right)^2 }  +\varepsilon .
$$
Taking limits as $\sigma \to \infty$, we have
\begin{align}
\lim_{\sigma\to\infty} \sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|
&\le \lim_{\sigma\to\infty}\sum_{n=1}^{n_0}  \frac{4}{ \left(e^{\pi\left(n-\frac{1}{2}\right)\sigma} - e^{-\pi\left(n-\frac{1}{2}\right)\sigma}\right)^2 }  +\varepsilon \\
&\le \varepsilon .
\end{align}
Since $\varepsilon $ is arbitrary, we see $$
\lim_{\sigma\to\infty} \sum_{n=1}^\infty \left|\frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}\right|=0,$$
which implies $$
\lim_{\sigma\to\infty} \sum_{n=1}^\infty \frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}
=0.$$
Similarly (we omit details, but just similarly) we have $$
\lim_{\sigma\to\infty} \sum_{n=-\infty}^0 \frac{1}{\cos ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}
=0$$
and $$
\lim_{\sigma\to\infty} \sum_{n=-\infty}^{\infty} \frac{1}{\sin ^2\left (\pi\left (n-\frac{1}{2}\right )\tau \right )}
=0,\quad
\lim_{\sigma\to\infty} \sum_{n=-\infty,\, n\ne 0}^\infty \frac{1}{\cos ^2\left (\pi n\tau \right )}
=0.$$
Thus we have $e_3-e_2\to 0,$ $e_1-e_2 \to \pi^2$.
</s>