Krdytkyu

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:

$$lim_{ato 0} frac{pi^2}{sin^2(pi a)} - frac{1}{a^2} .$$

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?

Follow your thought,
$$
lim_{ato 0}frac {pi^2}{sin^2(pi a)}-frac 1{a^2} = lim_{ato 0}frac {a^2pi^2 - sin^2(pi a)}{a^2 sin^2(pi a)} = lim_{ato 0}frac {(api - sin(pi a))(pi a + sin (pi a))}{a^2 sin^2(pi a)} = lim_{ato 0}frac {(api)^3/6 times 2pi a}{a^4pi^2 }= frac {pi^2} 3.
$$
However you should prove that
$$
lim_{ato 0}sum_{-infty}^{+infty} frac 1{(n+a)^2 } = sum_{-infty}^{+infty} lim_{ato 0}frac 1{(n+a)^2 } = sum_{-infty}^{+infty} frac 1{n^2}.
$$

Put $a=frac12$,

$$sum_{n=-infty}^infty frac1{(n+frac12)^2}=sum_{n=-infty}^infty frac4{(2n+1)^2}=pi^2impliessum_{n=-infty}^infty frac1{(2n+1)^2}=frac{pi^2}{4}.$$
Notice
$$sum_{n=-infty}^infty frac1{(2n+1)^2}=left(sum_{n=-infty}^{-1}+sum_{n=0}^inftyright)frac1{(2n+1)^2}=
2sum_{n=0}^infty frac1{(2n+1)^2}=frac{pi^2}{4}$$
implies
$$sum_{n=0}^infty frac1{(2n+1)^2}=frac{pi^2}8.$$

Let $S=sumlimits_{n=1}^inftyfrac 1{n^2}$, 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*}

Your idea of using Taylor expansions is good.

Let us compose the series around $a=0$
$$sin(pi a)=pi a-frac{pi ^3 a^3}{6}+frac{pi ^5 a^5}{120}+Oleft(a^7right)$$
$$sin^2(pi a)=pi ^2 a^2-frac{pi ^4 a^4}{3}+frac{2 pi ^6 a^6}{45}+Oleft(a^8right)$$
$$frac{pi^2}{sin^2(pi a)}=frac{pi^2}{pi ^2 a^2-frac{pi ^4 a^4}{3}+frac{2 pi ^6 a^6}{45}+Oleft(a^8right) }=frac{1}{a^2}+frac{pi ^2}{3}+frac{pi ^4 a^2}{15}+Oleft(a^4right)$$
$$frac{pi^2}{sin^2(pi a)} - frac{1}{a^2} =frac{pi ^2}{3}+frac{pi ^4 a^2}{15}+Oleft(a^4right)$$ which shows both the limit and also how it is approached.

Try with $a=frac 16$ which is "large". The exact value would be $4 pi ^2-36approx 3.47842$ while the expansion would give $frac{pi ^2}{3}+frac{pi ^4}{540}approx 3.47026$.