I am stuck on the following problem :
$\,\,\,\,$*Problem*$\quad$The residue of an entire function at $\infty$ is $0$.
Solution: True. This follows from the definition of the residue at $\infty$ together with the Cauchy-Goursat Theorem. Another way to see this is to take the Taylor expansion for $f$ at $0$ $\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n$ and replace $z$ with $1/z$ to get the Laurent expansion for $f(1/z)$ at $0$: $\displaystyle f(z)=\sum_{n=-\infty}^0a_{-n}z^n.$ Then we have that the residue at $\infty$ is given by the negative coefficient of the $1/z$ term of $$ \dfrac1{z^2}f\left(\dfrac1z\right)=\sum_{n=-\infty}^{-2}a_{-n-2}z^n, $$ which is clearly $0$.
I am having trouble to understand the last few lines (in italic) in the given solution. Can someone give lucid explanation? Thanks and regards to all.