Formula for $\zeta(2k)$ with Bernoulli numbers I want to derive the following formula for $\zeta(2k)$:
$$ \zeta(2k) = (-1)^{k - 1} 2^{2k - 1} \frac{B_{2k}}{(2k)!} \cdot \pi^{2k} $$
My approach was as follows: define $f(z) = 1/z^{2k}$, which is an even meromorphic function without poles or zeroes at the non-zero integers. Then I show that for every meromorphic function $f$ without poles or zeroes at the non-zero integers and for every non-zero integer $k$, the function
$$ g(z) = \frac{2 \pi i f(z)}{e^{2 \pi i z} - 1} $$
has residue $f(k)$ at $z = k$. Hence by the Residue theorem, we obtain
$$ \oint_{S_N} \frac{2\pi i f(z)}{e^{2 \pi i z} - 1} ~\text{d}z = 2\pi i \sum_{k = -N, ~ k \neq 0}^N f(k) = 4 \pi i \sum_{j = 1}^N f(j) $$
since $f$ was even. Here, $S_N$ is the square through the points $\pm (N + 1/2) \pm i (N + 1/2)$. Taking the limit of $N$ to infinity, we obtain
$$ 4 \pi i \cdot \zeta(2k) = \lim_{N \to \infty} \oint_{S_N} \frac{2\pi i f(z)}{e^{2 \pi i z} - 1} ~ \text{d}z $$
However, this is where things get weird. My teacher told me to use two things: (1) the fact that this limit is zero, and (2) the fact that the Bernoulli numbers are given by the formula
$$ \frac{z}{e^z - 1} = \sum_{n = 0}^{\infty} \frac{B_n}{n!} z^n $$
for $|z| < 2 \pi$. I flat out ignored (1) (because this would mean $\zeta(2k) = 0$, which we know isn't true), but using (2) would yield
$$ 4 \pi i \cdot \zeta(2k) = \lim_{N \to \infty} \oint_{S_N} \frac{f(z)}{z} \frac{2 \pi i z}{e^{2 \pi i z} - 1} ~\text{d}z = \lim_{N \to \infty} \oint_{S_N} \sum_{n = 0}^{\infty} \frac{f(z)}{z} \frac{B_n}{n!} (2 \pi)^n i^n z^n $$
But technically, I can't use (2), since $|2 \pi i z| = |2 \pi| \cdot |z|$ but $|z|$ can be bigger than 1 since we integrate over $S_N$. So in order to use (2), I probably have to use (1). But how?
Now if I just ignore the requirement $|2 \pi i z| < 2 \pi$, I arrive at
$$ 4 \pi i \cdot \zeta(2k) = \lim_{N \to \infty} \oint_{S_N} \sum_{n = 0}^{\infty} \frac{B_n}{n!} (2 \pi)^n i^n z^{n - 2k - 1} ~\text{d}z $$
Now I "know" I only have to consider $n = 2k$, because otherwise, the power $p$ in $z^p$ is different from $-1$, and then the line integral will be zero anyway (because $S_N$ is closed). However, I cannot justify interchanging the integral and the (infinite!) series. How can I do so? Because then, I arrive at
$$ 4 \pi i \cdot \zeta(2k) = \lim_{N \to \infty} \frac{B_{2k}}{(2k)!} (2 \pi)^{2k} i^{2k} \oint_{S_N} \frac{1}{z} ~\text{d}z = \lim_{N \to \infty} (-1)^k \frac{B_{2k}}{(2k)!} 2^{2k} \pi^{2k} \cdot 2 \pi i $$
which can then be written as
$$ \zeta(2k) = (-1)^k \frac{B_{2k}}{(2k)!} 2^{2k - 1} \pi^{2k} $$
which is almost what I wanted to show (the power of $-1$ should be $k - 1$ instead of $k$, but this is probably just some small error somewhere).
Any help is greatly appreciated.
NOTE: I am not looking for just any derivation of the formula for $\zeta(2k)$, since that can be found online. I am looking to derive the formula in this way, and in particular to understand why you have to show that the limit does indeed equal zero.
 A: After the line "Hence by the residue theorem" you forgot the residue at $0$.  So you should get $$4\pi i \zeta(2k) + 2\pi i \mathrm{Res}_{z = 0} g(z) = \lim_{N \to \infty} \oint_{S_n} g(z)\,dz.$$
Your teacher was right; you want to show that this limit is zero, and then you want to calculate the residue at $0$: I won't go through all the details, but the idea is: \begin{align*}
\mathrm{Res}_{z = 0}g(z) &= [z^{-1}] \frac{2\pi i }{z^{2k} (e^{2\pi i z} - 1)} \\
&=2\pi i [z^{2k}] \frac{z}{e^{2\pi i z} - 1}
\end{align*}
where $[z^a]$ denotes the coefficient of $z^a$ in the Laurent series.  You can find this explicitly in terms of the Bernoulli numbers now, and when the dust settles you'll get your formula.
A: This is how I'd write the solution

Let $$h(z) = \frac{z^{-2m}}{e^{ z}-1} = z^{-2m-1} \frac{z}{e^z-1}=z^{-2m-1} \sum_{k=0}^\infty \frac{B_k}{k!} z^k$$ then $$\int_{|z| = 2i\pi (N+1/2)} h(z)dz = \sum_{n=-N}^N \text{Res}(h(z),2i \pi n) = \frac{B_{2m}}{(2m)!}+ \sum_{n=-N, n \ne 0}^n (2i\pi n)^{-2m}$$
And since $|z| = 2i\pi (N+1/2) \implies h(z) = \mathcal{O}(N^{-2m})$  we obtain
$$0 = \lim_{N \to\infty}\int_{|z| = 2i\pi (N+1/2)} h(z)dz =\frac{B_{2m}}{(2m)!}+(2i\pi)^{-2m} 2 \zeta(2m)$$
$$\zeta(2m) = \frac{B_{2m}}{(2m)!}(-1)^{m-1}\frac12 (2\pi)^{2m}$$
