$\frac{z - 1}{z - 2} = \sum_{k \geq 0} \frac{1}{(z - 1)^k}$
by Hidenori
Proposition
Show that
\[\begin{align*} \frac{z - 1}{z - 2} = \sum_{k \geq 0}\frac{1}{(z - 1)^k} \end{align*}\]for $\abs{z - 1} > 1$.
Solution
Suppose $\abs{z - 1} > 1$.
\[\begin{align*} \frac{z - 1}{z - 2} &= \frac{z - 1}{(z - 1) - 1} \\ &= \frac{1}{1 - \frac{1}{z - 1}} \\ &= \sum_{k \geq 0} \frac{1}{(z - 1)^k}. \end{align*}\]Subscribe via RSS