Riemann Zeta Function at Even Integers/Proof 2

Proof
Let $k \in \N$.

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

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

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

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

By Fourier Series for Even Function over Symmetric Range:

We have:

Thus:

From the above:
 * $\zeta \left({2}\right) = \dfrac {\pi^2} 6$

which serves as our base case.

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

Then:

which completes the Induction Hypothesis.

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