<s>[INST] Determining the limit of this series $$ \sum_{n=0}^\infty  \frac{(-2)^n + 2^{3n}}{3^n4^n} = $$
$$ \sum_{n=0}^\infty \frac{(-1)^n2^n}{3^n4^n} + \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = $$
$$ \sum_{n=0}^\infty (-1)^n\frac{1}{6^n} + \sum_{n=0}^\infty \left(\frac{2}{3}\right)^n = $$
The second sum is a geometric series, and its limit is
$$ \frac{1}{1 - 2/3} = 3 $$
However, I am having trouble finding the limit of the first Sum. How else can I approach this question?
 [/INST] Both are geometric series:
$$
\sum_{n=0}^\infty \frac{(-2)^n}{3^n \cdot 4^n} = \sum_{n=0}^\infty  \left(\frac{-2}{3\cdot 4}\right)^n = \sum_{n=0}^\infty \left(\frac{-1}{6}\right)^n
$$
and
$$
\sum_{n=0}^\infty \frac{2^{3n}}{3^n\cdot 4^n} = 
\sum_{n=0}^\infty \frac{8^n}{3^n\cdot 4^n} =
\sum_{n=0}^\infty \left(\frac{8}{3\cdot 4}\right)^n =
\sum_{n=0}^\infty \left(\frac{2}{3}\right)^n
$$
</s>