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?
Best Answer
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 $$ \lim_{a\to 0}\sum_{-\infty}^{+\infty} \frac 1{(n+a)^2 } = \sum_{-\infty}^{+\infty} \lim_{a\to 0}\frac 1{(n+a)^2 } = \sum_{-\infty}^{+\infty} \frac 1{n^2}. $$