# [SOLVED]Sum from power series

#### dwsmith

##### Well-known member
Show $\sum\limits_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$ using the series for $(\pi\cot\pi z)'$ at $z = 0$

I know from class that $\sin\pi z = \pi z\prod\limits_{n\in\mathbb{Z} -\{0\}}\left[\left(1-\frac{z}{n}\right)e^{z/n}\right]$

So do I need to use that to rewrite cot as cosine over that product?

#### chisigma

##### Well-known member
A 'very nice' series expansion is...

$\displaystyle \pi \cot (\pi\ z)= \frac{1}{z}+2\ z\ \sum_{n=1}^{\infty} \frac{1}{z^{2}-n^{2}}$ (1)

From (1) You derive...

$\displaystyle \frac{{\pi\ \cot(\pi\ z)}-\frac{1}{z}}{2\ z}= \sum_{n=1}^{\infty} \frac{1}{z^{2}-n^{2}}$ (2)

Now compute the $\displaystyle \lim_{z \rightarrow 0}$ for both term of (2)...

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
A 'very nice' series expansion is...

$\frac{1}{z}+2\ z\ \sum_{n=1}^{\infty} \frac{1}{z^{2}-n^{2}}$ (1)
How did you come up with this piece?

#### chisigma

##### Well-known member
How did you come up with this piece?

$\displaystyle \sin (\pi\ z)= \pi\ z\ \prod_{n=1}^{\infty} (1-\frac{z^{2}}{n^{2}})$ (1)

... and first obtain...

$\displaystyle \ln \sin (\pi z) = \ln (\pi\ z) + \sum_{n=1}^{\infty} \ln (1-\frac{z^{2}}{n^{2}})$ (2)

Now if we derive (2) we obtain...

$\displaystyle \frac{d}{d z}\ \ln \sin (\pi\ z)= \pi\ \cot(\pi\ z)= \frac{1}{z} + 2\ z\ \sum_{n=1}^{\infty} \frac{1}{z^{2}-n^{2}}$ (3)

Kind regards

$\chi$ $\sigma$

#### dwsmith

##### Well-known member
$\displaystyle \frac{{\pi\ \cot(\pi\ z)}-\frac{1}{z}}{2\ z}= \sum_{n=1}^{\infty} \frac{1}{z^{2}-n^{2}}$ (2)

Now compute the $\displaystyle \lim_{z \rightarrow 0}$ for both term of (2)...

Kind regards

$\chi$ $\sigma$
I can't figure out how to take the limit of this function without using Mathematica.

Last edited:

#### chisigma

##### Well-known member
I can't figure out how to take the limit of this function without using Mathematica.
Take into account that is...

$\displaystyle \pi\ \cot (\pi\ z)= \frac{1}{z} - \frac{\pi^{2}}{3}\ z - \frac{\pi^{4}}{45}\ z^{3}-...$

Kind regards

$\chi$ $\sigma$