Changing index of summation

Question

How do you go from $\sum_{n=0}^{\infty} z^{-n-1}$ to $\sum_{n=-\infty}^{-1} z^{n}$ ?? It's really confusing.

Answer 1

$$ \sum_{n=0}^{\infty} \frac{1}{3} z^{-n-1} $$

Define $m=-n-1$. $n=0$ gives $m=-1$ and as $n$ goes up by $1$, m goes down by 1. So a $\sum_{n=0}^\infty$ becomes $\sum^{-1}_{-\infty}$ (Slight abuse of notation)

$$ \sum_{-\infty}^{-1} \frac{1}{3} z^{m} $$

Then rename $m$ to $n$.

Answer 2

Simply observe that for

$$\sum_{n=0}^{\infty} z^{-n-1} $$

when $n$ goes from $0 \to \infty$ the exponent $n-1$ goes form $-1 \to -\infty$ then the identity follows.