-1
$\begingroup$

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

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 $\Sigma$.

  • 0
    @AntonioVargas I couldn't $f$i$n$d that whilst I was searching thanks2012-11-29

1 Answers 1

3

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*}