Basel Problem/Proof 2

Proof
Let:


 * $\displaystyle P_k = x \prod_{n \mathop = 1}^k \paren {1 - \frac {x^2} {n^2 \pi^2} }$

We note that:

By Telescoping Series we find that the coefficient of $x^3$ in $P_k$ is given by:


 * $(1): \quad \displaystyle -\frac 1 {\pi^2} \sum_{i \mathop = 1}^k \frac 1 {i^2}$

From Euler Formula for Sine Function:


 * $\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$

So by taking the limit as $k \to \infty$ in $(1)$ and equating with the coefficient of $x^3$ in the Power Series Expansion for Sine Function, we can deduce:


 * $\displaystyle -\frac 1 {\pi^2} \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = -\frac 1 {3!}$

hence:


 * $\displaystyle \sum_{i \mathop = 1}^{\infty} \frac 1 {i^2} = \frac {\pi^2} 6$