Prove $\sum_{n=1}^{+\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$ from $\lim_{a \to 0+} \dfrac{\pi}{a} \dfrac{1 + e^{-2 \pi a}}{1 - e^{-2 \pi a}}$? Yes, I need to prove the solution to the Basel problem from a limit. The context is that we can combine the Fourier transform of $1/(n^2 + a^2)$ and the Poisson summation formula to show that (if $a 
> 0$) \begin{align} \sum_{n = -\infty}^{\infty} \dfrac{1}{n^2+a^2} &= \sum_{n = -\infty}^{-1} \dfrac{1}{n^2+a^2} + \sum_{n = 0}^{\infty} \dfrac{1}{n^2+a^2} \\&= \dfrac{\pi}{a} \sum_{n = -\infty}^{-1} e^{-2\pi a |n|} + \dfrac{\pi}{a} \sum_{n = 0}^{\infty} e^{-2\pi a n} 
\\&= \dfrac{\pi}{a} \left[\dfrac{r}{1-r} + \dfrac{1}{1-r} \right] \\&= \dfrac{\pi}{a} \dfrac{1+r}{1-r}, \text{ where } r = e^{-2 \pi a} \end{align}
Then I have to consider the limit $a \to 0^+$ to show $\displaystyle \sum_{n=1}^{+\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$.
My thoughts:
Since \begin{align} \dfrac{\pi}{a} \sum_{n = -\infty}^{-1} e^{-2\pi a |n|} &= \dfrac{\pi}{a} \sum_{n = 1}^{\infty} e^{-2\pi a n} \\&= \dfrac{\pi}{a} \dfrac{r}{1-r}\end{align} should I simply take $\displaystyle\lim _{a\to 0^+}\dfrac{\pi}{a} \dfrac{r}{1-r}$ ?
Thank you very much.

It turns out the correct way to approach this question is first recognizing \begin{equation} \sum_{n=-\infty}^{+\infty} \dfrac{1}{n^2 + a^2} = \sum_{n=-\infty}^{-1} \dfrac{1}{n^2 + a^2} + \sum_{n=1}^{+\infty} \dfrac{1}{n^2 + a^2} + \dfrac{1}{a^2} = 2\sum_{n=1}^{\infty} \dfrac{1}{n^2 + a^2} + \dfrac{1}{a^2} \end{equation} because $(-n)^2 = n^2$, so that \begin{equation} 2\sum_{n=1}^{\infty} \dfrac{1}{n^2 + a^2} = \dfrac{\pi}{a}\dfrac{1+\exp(-2\pi a)}{1-\exp(-2\pi a)} - \dfrac{1}{a^2},\end{equation} and by applying L'Hopital's rule around 3 times you would get the limit of RHS as $a \to 0^+$ to be $\dfrac{\pi^2}{3}$. The result follows.
