How can the sum of a geometric series be evaluated?

[thenewbosco]

i have the following series from 1 to infinity:
\sum(-1)^{n-1}*(\frac{4^n}{7^n})
how can i evaluate the sum of this?
thanks
i know the answer is 4/11 but i do not know how to get this.

 

[quasar987]

Look closer; it's just a geometrical serie.

What you might have missed: (-1)^{n-1}=(-1)^{n+1}.

 

[thenewbosco]

how does this fact help me find what the sum is equal to?

 

[quasar987]

Don't you know what the sum of a geometrical serie is?

http://mathworld.wolfram.com/GeometricSeries.html

 

[Pyrrhus]

It's like quasar said you need to rewrite into the form

\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}

where

a \neq 0