Let $z_1, z_2,\dots, z_k,\dots$ be all the roots of $e^z=z$.
Let $C_N$ be the square in the plane centered at the origin with siden parallel to the axis and each of length $2\pi N$.
Assume that $$\lim_{N\to \infty} \int\limits_{\displaystyle C_N}\frac{e^z-1}{z^2(e^z-z)}dz=0$$
Find $$\sum_{k=1}^{\infty}\frac{1}{z_k^2}$$
My solution attempt is too trivial. So I didn't write it here. Please, solve the question more explicitly. Thank you.
Let us call $f(z) = e^z - z$. Then the integral can be written
$$\int_{C_N} \frac{f'(z)}{z^2 f(z)}\,dz.$$
If $g$ is holomorphic on $\mathbb{C}\setminus D$, where $D$ is a closed discrete set in $\mathbb{C}$, then
$$\int_{C_N} g(z)\,dz$$
is $2\pi i$ times the sum of the residues of $g$ in the points of $D$ that are enclosed by $C_N$ (provided none of the points of $D$ lies on the contour $C_N$).
Here, the integrand $\dfrac{f'(z)}{z^2 f(z)}$ has singularities in the $z_k$ and in $0$. The residue in $z_k$ is $\dfrac{1}{z_k^2}$ since $f$ has only simple zeros, so
$$\int_{C_N} \frac{f'(z)}{z^2 f(z)}\,dz = 2\pi i \left(\operatorname{Res} \left(\frac{f'(z)}{z^2f(z)}; 0\right) + \sum_{z_k \in S_N} \frac{1}{z_k^2}\right).$$
Since the integral tends to $0$ for $N\to \infty$, we obtain
$$\sum_{k=1}^\infty \frac{1}{z_k^2} = -\operatorname{Res} \left(\frac{f'(z)}{z^2f(z)}; 0\right).$$
It remains to find that residue. Since $f(0) = 1$, we need the residue of $\dfrac{e^z-1}{z^2}$ in $0$, which is easily seen to be $1$.