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).
(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 $e_3-e_2\to 0,$ $e_1-e_2 \to \pi^2$.