Welcome to our community

Be a part of something great, join today!

Once very famous problem

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \sum_{k\geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}\)

Let us see how many different methods can we get (Cool)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
Jan 17, 2013
1,667
\(\displaystyle \sum_{k = -\infty}^{\infty} f(k) = - \sum_{i\geq 0} \text{Res}\, \left( \pi \cot(\pi z) f(z);z_i \right) \)

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.