Change of variables in Oppenheim example 2.5

Question

How do they do the change of variables from $$\sum_{k=-\infty}^{n} \rightarrow \sum_{l=-n}^{\infty}$$

For the top, you can write $n = m-l = m + k$, so as $k \rightarrow -\infty$ then $n\rightarrow -\infty$, but in their case the upper bound is $\infty$. I also am not sure how to start on the bottom part of the summation.

Answer 1

Somewhat more elaborated we have \begin{align*} \color{blue}{\sum_{k=-\infty}^{n}2^k}&=\lim_{m\to \infty}\sum_{k=-m}^n2^k\\ &=\lim_{m\to \infty}\sum_{k=-m}^n\left(\frac{1}{2}\right)^{-k}\tag{$2^k=\left(\frac{1}{2}\right)^{-k}$}\\ &=\lim_{m\to \infty}\sum_{k=-n}^m\left(\frac{1}{2}\right)^{k}\tag{$k\to -k$}\\ &\,\,\color{blue}{=\sum_{k=-n}^{\infty}\left(\frac{1}{2}\right)^{k}} \end{align*} according to the claim.

Without sum notation left-hand and right-hand sum from above are
\begin{align*} &\color{blue}{\cdots+2^{-1}+1+2+\cdots+2^{n-1}+2^n}\\ &\qquad\color{blue}{=\left(\frac{1}{2}\right)^{-n}+\left(\frac{1}{2}\right)^{-(n-1)}+\cdots+\left(\frac{1}{2}\right)^{-1}+1+\left(\frac{1}{2}\right)^{1}+\cdots} \end{align*}