Question

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

Answer 1

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

Answer 2

Yes

Answer 3

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

Answer 4

the common ratio is -1/4 and 1st term a = 3/4

then \[s _{\infty} = a/(1-r)\]

Answer 5

Thank you!

Answer 6

No problem. :D