What's with the lower bound of Laurent series?

Solution 1:

The general form of a Laurent series is $\sum_{n=-\infty}^\infty b_n z^n$, but this doesn't mean that each coefficient $b_n\neq0$. Using your example, for $|z|>2$, we have $$\frac{1}{z-2}=\frac{1}{z}\cdot\frac{1}{1-(2/z)}=\frac{1}{z}\sum_{n=0}^\infty \left(\frac{2}{z}\right)^n$$

Rewriting the series so that $b_n$ is more easily identified, we have $$\sum_{n=0}^\infty 2^n\left(\frac{1}{z}\right)^{n+1}=\sum_{n=1}^\infty 2^{n-1} z^{-n}$$

Observe that this series accounts for the negative integers, so another way to write this sum (though seemingly less aesthetically pleasing) is $$\sum_{-\infty}^{-1} 2^{-n-1}z^n$$

At this point, we can define $b_n$ in the following way: $$b_n=\begin{cases} 2^{-n-1}&n\leq-1\\ 0& n>-1\end{cases}$$ It follows from this definition for $b_n$ that $$\frac{1}{z-2}=\sum_{-\infty}^\infty b_nz^n,\quad|z|>2$$