<s>[INST] 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?
 [/INST] $$\sum\limits_{n=1}^\infty \frac{(-1)^n3^{n-2}}{4^n}=\sum\limits_{n=1}^\infty (-1)^n\frac{1}{9}\left(\frac{3}{4}\right)^n=\frac{1}{9}\sum\limits_{n=1}^\infty \left(-\frac{3}{4}\right)^n=
\frac{1}{9}\left(\frac{1}{1+3/4}-1\right)$$
</s>