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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?

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  • $\begingroup$ I am sure that this suggestion would work, but our assignment is that we have to use the first double infinite sum identity to show the second. $\endgroup$ – TunaBooties Nov 20 '18 at 3:46
  • $\begingroup$ It could help that $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$$ $\endgroup$ – clathratus Nov 20 '18 at 6:49
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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 $$ \lim_{a\to 0}\sum_{-\infty}^{+\infty} \frac 1{(n+a)^2 } = \sum_{-\infty}^{+\infty} \lim_{a\to 0}\frac 1{(n+a)^2 } = \sum_{-\infty}^{+\infty} \frac 1{n^2}. $$

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  • $\begingroup$ Thank you!! i didn't see the splitting the numerator up into the $(a^2 -b^2) = (a-b)(a+b)$ form! $\endgroup$ – TunaBooties Nov 20 '18 at 4:50
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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*}

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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 $\frac{\pi ^2}{3}+\frac{\pi ^4}{540}\approx 3.47026$.

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