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?

$$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n}=\sum_{n=1}^{\infty}\left(\frac{1}{3^n}+\frac{n^2}{2^n}\right),$$
where
$$\sum_{n=1}^{\infty}\frac{1}{3^n}=\frac{1}{1-\frac{1}{3}}-1=\frac{1}{2}$$
and, by ratio test,
$$\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}=\frac{1}{2}\left(1+\frac{1}{n}\right)^2 \to \frac{1}{2}<1 \;\; \text{ as } \;\; n \to \infty,$$
meaning that $\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ is absolutely convergent. Therefore, the whole series is convergent.
