Basel Problem/Proof 10

Proof
For $z \ne 0$, from Mittag-Leffler Expansion for Hyperbolic Cotangent Function, we have:


 * $\ds \frac 1 {2 z} \paren {\pi \map \coth {\pi z} - \frac 1 z} = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$

Consider the set:


 * $A = \set {z \in \C : \size z \le \dfrac 1 2}$.

Then for each $n \in \N$ and $z \in A$, we have:

For all $n \in \N$, we have:


 * $\ds n^2 - \frac 1 4 \ge \frac 1 2 n^2$

So that:


 * $\ds \frac 1 {n^2 - \frac 1 4} \le \frac 2 {n^2}$

From Basel Problem, we then have:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 2 {n^2} = \frac {\pi^2} 3 < \infty$

So by the Weierstrass M-Test, the series:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$

converges uniformly on $A$.

Then from Uniformly Convergent Series of Continuous Functions is Continuous, the function $f : A \to \C$ defined by:


 * $\ds \map f z = \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2}$

is continuous.

So:


 * $\ds \lim_{z \mathop \to 0} \sum_{n \mathop = 1}^\infty \frac 1 {z^2 + n^2} = \sum_{n \mathop = 1}^\infty \frac 1 {n^2}$

We can write:

So that:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \lim_{z \mathop \to 0} \paren {\frac {\pi z \paren {e^{2 \pi z} + 1} - e^{2 \pi z} + 1} {2 z^2 \paren {e^{2 \pi z} - 1} } }$

We have at $z = 0$:

and:


 * $2 z^2 \paren {e^{2 \pi z} + 1} = 0$

So by L'Hopital's Rule:

giving:


 * $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} = \frac {\pi^2} 6$