<s>[INST] Compute limit of $\sum_{n=0}^\infty \frac{(-2)^n+2^{3n}}{3^n\cdot4^n}$ 
I need help to compute the limit of the following series: $\sum_{n=0}^\infty \frac{(-2)^n+2^{3n}}{3^n\cdot4^n}$

What I got so far: 
It is probably easier to split the sum into the following:
$\sum_{n=0}^\infty \frac{(-2)^n+2^{3n}}{3^n\cdot4^n}$ = $\sum_{n=0}^\infty \frac{(-2)^n}{(3\cdot4)^n} + \sum_{n=0}^\infty \frac{8^{n}}{(3\cdot4)^n}=$
$\sum_{n=0}^\infty (\frac{-1}{6})^n + \sum_{n=0}^\infty (\frac{8}{12})^n= \sum_{n=0}^\infty (-1)^n (\frac{1}{6})^n + \sum_{n=0}^\infty (\frac{4}{6})^n  = \sum_{n=0}^\infty (-1)^n (\frac{1}{6})^n + \frac{1}{1-\frac{4}{6}} = \sum_{n=0}^\infty (-1)^n (\frac{1}{6})^n + 3$
But here im stuck: I don't know how to deal with the alternating series that looks similar to another geometric series. Help to get further is very appreciated!
 [/INST] use the geometric series
$${\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\quad {\text{ for }}|x|<1\!$$
and let $x\rightarrow -x$ to get
$${\frac {1}{1+x}}=\sum _{n=0}^{\infty }(-1)^n(x)^{n}\quad {\text{ for }}|x|<1\!$$
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