Riemann Zeta Function at Even Integers/Proof 2

Proof
Let $k \in \N$

Let $S\left(x\right)$ be equal $x^{2k}$ on $\left[-\pi,\pi\right]$ and be periodic with period $2\pi$

Let $\displaystyle I\left(2m, n \right) = \int_0^\pi x^{2m} \cos\left( n x\right) \rd x$

Let $\displaystyle A\left(2m, n\right) = \frac {\pi^{2m-1} \left(-1\right)^{n} 2m} {n^2}$

Let $\displaystyle B\left(2m, n\right) =-\frac{2m\left(2m-1\right)}{n^2}$

By Fourier Series for Even Function over Symmetric Range:

We have:

Thus:

From the above, We have that:
 * $\zeta\left(2\right) = \frac {\pi^{2} } {6} $

which serves as our base case.

Assume, for induction, that for $1\le m \le k-1$
 * $ \displaystyle \zeta\left(2m\right) = \left({-1}\right)^{m + 1} \dfrac {B_{2 m} 2^{2 m - 1} \pi^{2 m} } {\left({2 m}\right)!}$

Then:

Completing the Induction Hypothesis.

Thus, for all $n \ge1$
 * $\displaystyle \zeta \left({2 n}\right) = \left({-1}\right)^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\left({2 n}\right)!}$