Q:

Calculate $\sum_{n=0}^{\infty} z^{n-2}/5^{n-1}$ for $0<|z|<5$

Possible Duplicate:
Complex series: $\sum_{n=0}^\infty\left( z^{n-2}/5^{n+1}\right)$ for $0 &amp;lt; |z| &amp;lt; 5$ 

I don't even know where to start. I can't think of any formulas or anything. I just know that the answer needs to be without a $\Sigma$.

A:

You surely know the geometric series $\sum_{n=0}^\infty q^n = \frac 1{1-q}$, which holds true for $|q| < 1$. You are given $0 < |z| < 5$, that is $\frac{|z|}5 < 1$, and your term looks alike. So let's give it a try. We let $q := \frac z5$, then 
\begin{align*}
  \sum_{n=0}^\infty \frac{z^{n-2}}{5^{n-1}} &= \frac{5}{z^2} \sum_{n=0}^\infty \frac{z^n}{5^n}\\
   &= \frac 5{z^2}\sum_{n=0}^\infty \left(\frac z5\right)^n\\ 
   &= \frac 5{z^2} \cdot \frac 1{1 - \frac z5}\\
   &= \frac {25}{z^2(5 - z)}.
\end{align*}

