Proving ∏ with summation and/or Lhopitals?

[LearninDaMath]

Homework Statement

Someone in school was showing me this proof or problem that, I believe, proves or yields π via this limit: Lim x-->0 of [itex]\frac{xπcot(πx)}{x}[/itex]-[itex]\frac{1}{x}[/itex] = tan(0) = 0

And that this somehow related to a summation [itex]\sum1/k^{2}[/itex] as the sum goes from 1 to ∞.

I don't recall any of the details about the proof or what the proof is proving (but I think it has something to do with proving that that π=3.14 as the sum gets closer to infinity.)

Does anything like this look or sound familiar to any known proof?

 

[Dustinsfl]

Homework Statement

Someone in school was showing me this proof or problem that, I believe, proves or yields π via this limit: Lim x-->0 of [itex]\frac{xπcot(πx)}{x}[/itex]-[itex]\frac{1}{x}[/itex] = tan(0) = 0

And that this somehow related to a summation [itex]\sum1/k^{2}[/itex] as the sum goes from 1 to ∞.

I don't recall any of the details about the proof or what the proof is proving (but I think it has something to do with proving that that π=3.14 as the sum gets closer to infinity.)

Does anything like this look or sound familiar to any known proof?

There seems to be some confusion with what you have.

First ##\sum\limits_{n=1}^{\infty}\frac{1}{n^2}=\frac{ \pi ^2}{6}##.

Here is the proof. I will be using the series for ##(\pi\cot\pi z)'## at ##z=0##

We have that ##\sin\pi z = \pi z\prod\limits_{n = 1}^{\infty}\left(1 - \frac{z^2}{n^2}\right)## which is an entire function with simple zeros exactly at the integers.
We can differentiate termwise by uniform convergence.
So by logarithmic differentiation, we obtain a series for ##\pi\cot\pi z##.
$$
\frac{d}{dz}\ln\left(\sin\pi z\right) = \pi\cot\pi z = \frac{1}{z} - 2z\sum_{n = 1}^{\infty}\frac{1}{n^2 - z^2}
$$
So we have that ##-\sum\limits_{n = 1}^{\infty}\dfrac{1}{n^2 - z^2} = \dfrac{\pi\cot\pi z - \frac{1}{z}}{2z}##.
In order to observe that the limit as ##z## goes to 0 works, we need to look at the power series of
$$
\pi\cot\pi z = \frac{1}{z} - \frac{\pi^2}{3}z - \frac{\pi^4}{45}z^3 -\cdots.
$$
So by taking the Taylor series expansion of ##\cot\pi z## and multiplying by ##\pi##, we have
\begin{align}
\frac{\pi\cot\pi z - \frac{1}{z}}{2z} &= \frac{- \frac{\pi^2}{3}z - \frac{\pi^4}{45}z^3 -\cdots}{2z}\notag\\
&= -\frac{\pi^2}{6} - \frac{\pi^4}{90}z^2 -\cdots.\notag
\end{align}
At this point, the singularity at 0 has been removed.
So
\begin{align}
\lim_{z\to 0}-\sum_{n = 1}^{\infty}\frac{1}{n^2 - z^2} = \lim_{z\to 0}\frac{\pi\cot\pi z - \frac{1}{z}}{2z}\notag\\
-\sum_{n = 1}^{\infty}\frac{1}{n^2} = \lim_{z\to 0}\left[\frac{\pi\cot\pi z - \frac{1}{z}}{2z}\right]\notag\\
\sum_{n = 1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.\notag
\end{align}