Question

Determine whether the following series converges or diverges. If it converges, calculate the sum

lower limit = 1
upper limit = infinity
((-3)^(n-1))(4^(-n))

Answer 1

\[\sum_{n=1}^{\infty} (-3)^{n-1}4^{-n}\]

Answer 2

it is a geometric sum

it converges

Answer 3

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

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

\[=\frac{1}{4}\sum_{n=1}^{\infty} \left(-\frac{3}{4}\right)^{n-1}\]