Showing that the limit of contour integrals is zero I want to prove that

$$\lim\limits_{N \to \infty}{\oint_{C_N}{\frac{z}{\exp(z)-1}}\cdot\frac{dz}{z^{2\cdot k+1}}}=0,$$
  where $C_N=\{z\in \mathbb C : |z|=2\pi(N+\frac{1}{2}) \}$ and $k\in \mathbb N$ is fixed.

I understand that this holds:
$$|\oint_{C_N}{\frac{z}{exp(z)-1}}\cdot\frac{dz}{z^{2\cdot k+1}}|\le\oint_{C_N}{|\frac{1}{\exp(z)-1}}\cdot\frac{1}{z^{2\cdot k}}|\le2\pi \cdot 2\pi(N+\frac{1}{2}) \cdot \frac{1}{(2\pi(N+\frac{1}{2}))^{2\cdot k}} \cdot \max\{\frac{1}{|\exp(z)-1|}|z\in C_N\} $$
and I know that $\frac{1}{|\exp(z)-1|}$ has a non zero denominator on $C_N$, and a maximum because $C_N$ is compact.
But I struggle to find an upper bound for this term.
Could anyone help me?  
 A: Let $\mu=2\pi (N+\frac12)$. We have: 
$$
z=\mu  e^{i\theta}=\mu\left[\cos (\theta) + i \sin (\theta)\right] 
$$
and
$$
e^z-1=e^{\mu \cos(\theta)}[\cos(\mu \sin(\theta))+i\sin (\mu \sin(\theta))]-1=e^{\mu \cos(\theta)}\cos(\mu \sin(\theta))-1+i\sin (\mu \sin(\theta))=F(\theta)
$$
Let $g(\theta)=|F(\theta)|^2$
$$
g(\theta)=|F(\theta)|^2=[e^{\mu \cos(\theta)}\cos(\mu \sin(\theta))-1]^2+[\sin (\mu \sin(\theta))]^2
$$
with
$$
\min_{\theta \in [0,2\pi]} g(\theta)=g(\pi)
$$
$$
z=\mu e^{i\pi}=-\mu
$$
Thus
$$\max_{z \,\in \, C_N} \frac{1}{|e^z-1|}=\frac{1}{|e^{-\mu}-1|}=\frac{1}{|e^{-2\pi(N+\frac12)}-1|}
$$
Given that, we have:
$$
{\oint_{C_N}{\frac{z}{\exp(z)-1}}\cdot\frac{dz}{z^{2\cdot k+1}}} \le\oint_{C_N}{|\frac{1}{\exp(z)-1}}\cdot\frac{1}{z^{2\cdot k}}|\le2\pi \cdot 2\pi(N+\frac{1}{2}) \cdot \frac{1}{(2\pi(N+\frac{1}{2}))^{2\cdot k}} \cdot \frac{1}{|e^{-2\pi(N+\frac12)}-1|}=2\pi \cdot   \frac{1}{(2\pi(N+\frac{1}{2}))^{2k-1}} \cdot \frac{1}{|e^{-2\pi(N+\frac12)}-1|} 
$$
and as $N\to \infty$:
$$\lim\limits_{N \to \infty}{\oint_{C_N}{\frac{z}{\exp(z)-1}}\cdot\frac{dz}{z^{2\cdot k+1}}}=\lim_{N \to \infty}2\pi \cdot   \frac{1}{(2\pi(N+\frac{1}{2}))^{2k-1}} \cdot \frac{1}{|e^{-2\pi(N+\frac12)}-1|}=2\pi \cdot 0 \cdot 1=0$$

It was assumed that $k \in \mathbb{N}\setminus \{0\}$
