Question

Determine if the series is convergent or divergent, if it's convergent, find the sum:

Answer 1

\[\sum_{n=1}^{\infty}\frac{ (-3)^{n-1} }{ 4^n }\] can u guys help me out ?

Answer 2

Recall that for \(|r|<1\),
\[\sum_{n=0}^\infty ar^n=\sum_{n=1}^\infty ar^{n-1}=\frac{a}{1-r}\]
In this case, \(r=-\dfrac{3}{4}\), so the series indeed converges.
\[\frac{(-3)^{n-1}}{4^n}=\frac{(-3)^{n-1}}{4\times4^{n-1}}=\frac{1}{4}\left(-\frac{3}{4}\right)^{n-1}\]
so \(a=\dfrac{1}{4}\).

Answer 3

how do you get to express the 4^n to (4x4^n)?

Answer 4

I meant 4x4^(n-1)

Answer 5

I'm guessing like this