$$\frac{e^{z^2}}{z^{2n+1}}$$ Am I right that limit as z approaches infinity does not exist? So its residue at infinity is equal to $c_{-1}$ of Laurent series. How am I supposed to get Laurent series of this function? Where is it centered? What range?
So $$e^{z^2}z^{-2n-1}$$ should somehow get to $$\sum_{k=-\infty}^{+\infty}{A_k(z-z_0)^k}$$
On
Strictly speaking, one finds residues of differentials, not of functions. Here the differential is $$\alpha=\frac{\exp(z^2)}{z^{2n+1}}\,dz$$
To find the residue at $\infty$, set $z=1/w$ and expand in powers of $w$. The coefficient of $w^{-1}\,dw$ is the residue.
Applying this to $\alpha$ gives $$\alpha=w^{2n+1}\exp(1/w^2)\left(-\frac{dw}{w^2}\right) =-\left(\sum_{k=0}^\infty\frac{w^{2n-1}}{k!w^{2k}}\right)\,dw.$$ The coefficient of $w^{-1}\,dw$ is $-1/n!$, and that is the residue of the differential $\alpha$ at $\infty$.
The residue at infinity of an analytic function $f$ is the residue at $0$ of $\frac{-1}{z^2}f\left(\frac1z\right)0$. In the case of the function that you mentioned, it's the residue at $0$ of$$\frac{-1}{z^2}\times z^{2n+1}e^{\frac1{z^2}},$$which is easy to compute, since$$e^{\frac1{z^2}}=1+\frac1{z^2}+\frac1{2!z^4}+\cdots$$