((-4)^(n-1))/5^n
Follow Math Help Forum on Facebook and Google+
Are you trying to determine convergence? $\displaystyle \sum \frac{(-4)^{n-1}}{5^n} \ = \ \sum \frac{(-1)^{n-1} 4^{n-1}}{5^n} \cdot {\color{red}\frac{4}{4}} \ = \ \frac{1}{4}\sum (-1)^{n-1}\left(\frac{4}{5}\right)^n$ Use the alternating series test.
oops sorry i meant to ask what was the sum of that equation.
Originally Posted by amiv4 oops sorry i meant to ask what was the sum of that $\displaystyle \color{red}\text{series}$. $\displaystyle \forall{x}\backepsilon|x|<1~\sum_{n=0}^{\infty}x^n =\frac{1}{1-x}$ So you have $\displaystyle \frac{-1}{4}\sum_{n=0}^{\infty}\left(\frac{4}{5}\right)^n$ which fits the bill, so the sum is $\displaystyle \frac{-1}{4}\cdot\frac{1}{1+\frac{4}{5}}$
View Tag Cloud