1. ## Series

((-4)^(n-1))/5^n

2. Are you trying to determine convergence?

$\sum \frac{(-4)^{n-1}}{5^n} \ = \ \sum \frac{(-1)^{n-1} 4^{n-1}}{5^n} \cdot {\color{red}\frac{4}{4}} \ = \ \frac{1}{4}\sum (-1)^{n-1}\left(\frac{4}{5}\right)^n$

Use the alternating series test.

3. oops sorry i meant to ask what was the sum of that equation.

4. Originally Posted by amiv4
oops sorry i meant to ask what was the sum of that $\color{red}\text{series}$.
$\forall{x}\backepsilon|x|<1~\sum_{n=0}^{\infty}x^n =\frac{1}{1-x}$

So you have $\frac{-1}{4}\sum_{n=0}^{\infty}\left(\frac{4}{5}\right)^n$ which fits the bill, so the sum is $\frac{-1}{4}\cdot\frac{1}{1+\frac{4}{5}}$