# Once very famous problem

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \sum_{k\geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}$$

Let us see how many different methods can we get

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Euler used the Bernoulli numbers to solve the problem

But there are other ways

Fourier series , Complex analysis

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \sum_{k = -\infty}^{\infty} f(k) = - \sum_{i\geq 0} \text{Res}\, \left( \pi \cot(\pi z) f(z);z_i \right)$$

This requires the function to be analytic on the real axis but $$\displaystyle \frac{1}{k^2}$$ has a pole of order $$\displaystyle 2$$ at the origin .

So we can adjust the theorem to solve for poles on the real axis

$$\displaystyle \sum_{k\leq -1} \frac{1}{k^2} +\sum_{k \geq 1} \frac{1}{k^2}= - \text{Res}\, \left(\frac{\pi \cot(\pi z)}{z^2};0 \right)$$

$$\displaystyle \sum_{k\geq 1} \frac{1}{k^2} +\sum_{k\geq 1} \frac{1}{k^2}= \frac{\pi^2}{3}$$

$$\displaystyle \sum_{k\geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}$$

I deleted the proof of the modification of the theorem above . If someone is interested I will try to post it.