Sum from 1 to infinity?

Question

Ok this is really annoying me:
The question is

(sigma) where n=1 to infinity; is equal to:

[(-3)^(n-1)]/[8^n]

I understand it is an alternating series and that it is convergent.

But, I do not know how to go about figuring out what it converges to.

I don't know how to go about doing this.

Help please...

hmmm thanks but no thanks - to anyone else reading this question this is simply a geometric sequence - the sum to infinity is equal to:

the first term/ 1-r

Answer 1

[(-3)^n * (-3)^(-1)] / (8^n)
[(-1 * 3)^n ) / [3 * (8^n)]
(-1/3) * (3/8)^n * (-1)^n

(3/8)^n -----> 0

Thus, the series converges by alternating series test.

Answer 2

[(-3)^(n-1)]/[8^n] = [(-3)^n (-3)^(-1)]/[8^n] = (-1/3)[(-3)^n]/[8^n] = (-1/3)[(-3/8)^n]