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?
 A: Follow your thought, 
$$
\lim_{a\to 0}\frac {\pi^2}{\sin^2(\pi a)}-\frac 1{a^2} = \lim_{a\to 0}\frac {a^2\pi^2 - \sin^2(\pi a)}{a^2 \sin^2(\pi a)} = \lim_{a\to 0}\frac {(a\pi - \sin(\pi a))(\pi a + \sin (\pi a))}{a^2 \sin^2(\pi a)} = \lim_{a\to 0}\frac {(a\pi)^3/6 \times 2\pi a}{a^4\pi^2  }= \frac {\pi^2} 3. 
$$
However you should prove that
$$
\lim_{a\to 0}\sum_{-\infty}^{+\infty} \frac 1{(n+a)^2 } = \sum_{-\infty}^{+\infty} \lim_{a\to 0}\frac 1{(n+a)^2 } = \sum_{-\infty}^{+\infty} \frac 1{n^2}.
$$
A: Put $a=\frac12$,
$$\sum_{n=-\infty}^\infty \frac1{(n+\frac12)^2}=\sum_{n=-\infty}^\infty \frac4{(2n+1)^2}=\pi^2\implies\sum_{n=-\infty}^\infty \frac1{(2n+1)^2}=\frac{\pi^2}{4}.$$
Notice
$$\sum_{n=-\infty}^\infty \frac1{(2n+1)^2}=\left(\sum_{n=-\infty}^{-1}+\sum_{n=0}^\infty\right)\frac1{(2n+1)^2}=
2\sum_{n=0}^\infty \frac1{(2n+1)^2}=\frac{\pi^2}{4}$$
implies
$$\sum_{n=0}^\infty \frac1{(2n+1)^2}=\frac{\pi^2}8.$$
Let $S=\sum\limits_{n=1}^\infty\frac 1{n^2}$, separate $S$ into odd and even terms,
\begin{align*}
\underbrace{\sum_{n=1}^\infty \frac1{n^2}}_{=S}&=\sum_{n=0}^\infty \frac1{(2n+1)^2}+\underbrace{\sum_{n=1}^\infty \frac1{(2n)^2}}_{=\frac14 S}\\
S&=\frac{\pi^2}8+\frac 14 S\\
S&=\frac{\pi^2}{6}.
\end{align*}
A: 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 $\frac{\pi ^2}{3}+\frac{\pi ^4}{540}\approx 3.47026$.
