Double Infinite sum of $1/n^2$ I am trying to use an identity we showed on our homework:  
$$ \sum_{-\infty}^{\infty} \frac{1}{(n+a)^2}  = \frac{\pi^2}{\sin^2(\pi a)} $$
to show that $$ \sum_{1}^{\infty} \frac{1}{n^2}  = \frac{\pi^2}{6}.$$
I have broken the first double infinite sum into the sum from $-\infty$ to $1$ plus the the $0^{th}$ term, plus the sum from $1$ to $+\infty$ and then I want to take the limit of $a$ going to $0$. 
This results in taking the limit of the following:
$$\lim_{a\to 0} \frac{\pi^2}{\sin^2(\pi a)} - \frac{1}{a^2}   .$$
Which I know should result in $\frac{\pi^3}{3}$ from Wolfram Alpha, as desired, but I am struggling with showing it analytically. 
My idea was to try and find the Maclaurien Expansion of $\sin^2(\pi a)$, but then taking that series to the exponent of negative 1 since it is in the denominator is causing issues. 
Is there a trick I am not seeing or a possible better way to use the above property to show the other infinite sum? 
This question is also for a complex analysis class, so perhaps there is a way to use complex Laurent or power series?

Your idea of using Taylor expansions is  good.
Let us compose the series around $a=0$
$$\sin(\pi a)=\pi  a-\frac{\pi ^3 a^3}{6}+\frac{\pi ^5 a^5}{120}+O\left(a^7\right)$$
$$\sin^2(\pi a)=\pi ^2 a^2-\frac{\pi ^4 a^4}{3}+\frac{2 \pi ^6 a^6}{45}+O\left(a^8\right)$$
$$\frac{\pi^2}{\sin^2(\pi a)}=\frac{\pi^2}{\pi ^2 a^2-\frac{\pi ^4 a^4}{3}+\frac{2 \pi ^6 a^6}{45}+O\left(a^8\right) }=\frac{1}{a^2}+\frac{\pi ^2}{3}+\frac{\pi ^4 a^2}{15}+O\left(a^4\right)$$
$$\frac{\pi^2}{\sin^2(\pi a)} - \frac{1}{a^2} =\frac{\pi ^2}{3}+\frac{\pi ^4 a^2}{15}+O\left(a^4\right)$$ which shows both the limit and also how it is approached.
Try with $a=\frac 16$ which is "large". The exact value would be $4 \pi ^2-36\approx 3.47842$ while the expansion would give $\frac{\pi ^2}{3}+\frac{\pi ^4}{540}\approx 3.47026$.
