Is the coefficient of the series $-\frac{x}{2}\cot(\pi x)= \zeta(2n)=\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi )^{2n}}{2(2n)!}$
The Maclaurin series formula for $\cot(x)$ is $\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}x^{2n-1}}{(2n)!}$
$$-\dfrac{x}{2}\cot(\pi x)= \displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^{2n-1}}{2(2n)!}$$
If this is so, why does wiki gave the generating function:
"The values of the zeta function at non-negative even integers have the generating function:"
$$\displaystyle\sum_{n=0}^{+\infty}\zeta(2n)x^{2n}=\dfrac{\pi x}{2}\cot(\pi x)=-\dfrac{1}{2}+\dfrac{\pi^2}{6}x^2+\dfrac{\pi^4}{90}x^4+\dfrac{\pi^6}{945}x^6...$$
Why do we need an extra $\pi$
Wiki proof also employ the function $-\pi x\cot(\pi x)$ for its proof:
https://proofwiki.org/wiki/Riemann_Zeta_Function_at_Even_Integers
Why not use this function, which is easier. In fact, after deriving the formula for Maclaurin series of $\cot(x)$, I have come up with this function, but I thought I was wrong.