Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?

Hint:  $\lim_{n \rightarrow \infty} \dfrac {n^4}{2^n} = 0$
$\implies $ After a particular value of $m \in \mathbb N :  n^4 << 2^n ~\forall~ n \ge m$ or 
$\dfrac {n^4}{2^n}<1  ~\forall~ n \ge m$.
Hence, $\dfrac {n^2}{2^n} < \dfrac {1}{n^2} \implies \sum \dfrac {n^2}{2^n} <  \sum \dfrac {1}{n^2}    $
and $\sum \dfrac {1}{n^2}    $ is a convergent series.
Hence, $\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n} = \sum_{n=1}^{\infty} \dfrac {1}{3^n} + \dfrac {n^2}{2^n}$ is convergent
