Laurent series and residue.

Question

Why is $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, using the identity ($m=-n$)? Why shouldn't the exponent be equal to $2m-1$? Also, why is the following correct - $\sum_{n=0}^\infty (1/n!)(1/z)^n = \sum_{m=-\infty}^0 z^{-m}/(-m)!$? Why shouldn't the exponent of $z$ be equal to $m$? And last, why is the $a_{-1}$ term of this Laurent series (i.e. the residue) equal to $1$ at $z = 0$?

Answer 1

Indeed, the exponent is $2m-1$ and not $2m+1$. Respect to the other question it is evident that the exponent of $z$ is $m$, otherwise the equality only holds for $z=0$. In this case the Laurent series at $z=0$ is $$\sum_{n=0}^{\infty}\frac{1}{n!}\frac{1}{z^n}=a_{0}+\frac{a_1}{z}+\frac{a_2}{z^2}+\cdots$$ where the term $a_1$ is the residue by definition, which in this case is 1.

Answer 2

\begin{align} \sum_{n=0}^\infty (1/w)^{2n+1} &= \sum_{n=0}^\infty w^{-2n-1}\\ &=\sum_{n=\infty}^0 w^{-2n-1}\\ &=\sum_{-m=\infty}^0 w^{-2n-1}\\ &=\sum_{m=-\infty}^0 w^{2m+1}\\ \end{align}