# Convergence of series

1. Oct 12, 2005

### 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.

2. Oct 12, 2005

### quasar987

Look closer; it's just a geometrical serie.

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

3. Oct 12, 2005

### thenewbosco

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

4. Oct 13, 2005

### quasar987

5. Oct 13, 2005

### Pyrrhus

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

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

where

$$a \neq 0$$