Working with exponent on series Hi have this sequence:
$$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}$$
I understand that this is a Geometric series so this is what I've made to get the sum.
$$\sum\limits_{n=1}^\infty (-1)^n\frac{3^{n}\cdot 3^{-2}}{4^n}$$
$$\sum\limits_{n=1}^\infty (-1)^n\cdot 3^{-2}{(\frac{3}{4})}^n$$
So $a= (-1)^n\cdot 3^{-2}$ and $r=\frac{3}{4}$ and the sum is given by
$$(-1)^n\cdot 3^{-2}\cdot \frac{1}{1-\frac{3}{4}}$$
Solving this I'm getting the result as $\frac{4}{9}$ witch I know Is incorrect because WolframAlpha is giving me another result.
So were am I making the mistake?
 A: The objective here is to transform your sum into a sum of the form: 
$$\sum_{n=1}^\infty ar^{n-1}$$
$$\text{Transformation: }\quad\quad\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-3)^{n-1}}{4^{n-1}} = \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\left(\frac{-3}{4}\right)^{n-1}$$
Hence $a = -\dfrac{1}{12}$ and $r = -\dfrac{3}{4}.\quad$ Now use the fact that 
$$\sum_{n=1}^\infty ar^{n-1} = \dfrac{a}{1 - r} = -\left(\frac{1}{12}\right)\cdot \left(\frac{1}{1 - (-\frac{3}{4})}\right)$$
Simpilfy, and then you are done!
A: $$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}=\sum\limits_{n=1}^\infty (-1)^n\frac{1}{9}\left(\frac{3}{4}\right)^n=\frac{1}{9}\sum\limits_{n=1}^\infty \left(-\frac{3}{4}\right)^n=
\frac{1}{9}\left(\frac{1}{1+3/4}-1\right)$$
A: You have
$$\begin{align}
\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n} 
&= \sum_{n=1}^{\infty} \frac{(-1)(-1)^{n-1}\frac{1}{3}3^{n-1}}{4\cdot 4^{n-1}} \\ 
&= \sum_{n=1}^{\infty} \frac{-1}{4\cdot 3}\frac{(-1)^{n-1}3^{n-1}}{4^{n-1}}\\
&= \sum_{n=1}^{\infty} \frac{-1}{12}\left(\frac{-3}{4}\right)^{n-1}
\end{align}$$
So you have $a = \frac{-1}{12}$ and $r = \frac{-3}{4}$.
Note the key thing here that both $a$ and $r$ are constants/numbers. They do not depend on $n$. The idea is that you rewrite your series so that it is of exactly the form
$$
\sum_{n=1}^{\infty} a r^{n-1}
$$
where again $a$ and $r$ are constants/numbers.
