## anonymous 4 years ago Find the sum of the Series: Sigma (-1)^(n-1) (3/4^n) from n=1 to infinity

1. blockcolder

You mean this: $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{3}{4^n}$

2. anonymous

Yes

3. blockcolder

This is an alternating geometric series with r=-1/4 and a=3/4. Thus, the sum is $\frac{\frac{3}{4}}{1+\frac{1}{4}}=\frac{3}{5}$.

4. campbell_st

the common ratio is -1/4 and 1st term a = 3/4 then $s _{\infty} = a/(1-r)$

5. anonymous

Thank you!

6. blockcolder

No problem. :D