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?
 A: You can calculate the exact value of that sum. Since $\sum_{k\geq 0}x^k=\frac{1}{1-x}$ if |x|<1, uniformly in $k$, you can differentiate this expression to get $\sum_{k\geq 0}x^k=\frac{x}{(1-x)^2}$, differentiate to get an expression for $\sum_{k\geq 0}k^2x^k$, wherer |x|<1. In your case, $x=1/2$. The other portion is a geometric progression. 
A: We might compare and then apply either the root or ratio test, if these are your preferred tests. That is
$$
\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n} < \sum_{n = 1}^\infty \frac{n^23^n + n^23^n}{6^n} = 2\sum_{n = 1}^\infty \frac{n^2 3^n}{6^n} = 2\sum_{n = 1}^\infty \frac{n^2}{3^n},
$$
and you can now use either the root test or the ratio test (or perhaps a variety of other methods) to determine the convergence of the series. 
Thematically, the exponential growth in the denominator is so much larger than the polynomial growth in the numerator of the original series that we can vastly overestimate like this and be fine.
A: 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
A: $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n}=\sum_{n=1}^{\infty}\left(\frac{1}{3^n}+\frac{n^2}{2^n}\right),$$
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.
A: $$\sum\limits_{n=1}^{\infty}\frac{2^n+n^2 3^n}{6^n}$$
I wouldn't recommend the root test for this series. However, here are the steps 
$$r=\limsup\limits_{n\to\infty}\sqrt[n]{\left|\frac{2^n+n^2 3^n}{6^n}\right|}$$
$$=\limsup\limits_{n\to\infty}\frac{\sqrt[n]{2^n+n^2 3^n}}{\sqrt[n]{6^n}}$$
$$=\frac16\limsup\limits_{n\to\infty}\left(2^n+n^2 3^n\right)^{\frac{1}{n}}$$
$$=\frac16\limsup\limits_{n\to\infty}\exp\left(\log\left(2^n+n^2 3^n\right)^{\frac{1}{n}}\right)$$
$$=\frac16\exp\left(\limsup\limits_{n\to\infty}\frac{1}{n}\log\left(2^n+n^2 3^n\right)\right)$$
$$=\frac16\exp\left(\limsup\limits_{n\to\infty}\frac{2^n\log 2+2n3^n+n^2 3^n\log 3}{2^n+n^2 3^n}\right)$$
$$=\frac16\exp\left(\limsup\limits_{n\to\infty}\frac{\frac{2^n\log 2}{n^2 3^n}+\frac{2}{n}+\log 3}{\frac{2^n}{n^2 3^n}+1}\right)$$
$$=\frac16\exp\left(\log 3\right)=\frac36=\frac12$$
Note that $r\lt 1$, therefore
$$\sum\limits_{n=1}^{\infty}\frac{2^n+n^2 3^n}{6^n}=\mbox{convergent} $$
