Is the series $\sum_{n=1}^\infty \frac{3^n n^2}{n!}$ convergent Is the following series convergent or divergent?
$$\sum_{n=1}^\infty \frac{3^n n^2}{n!}$$
I used root test to check but I am stuck at
$$\lim\limits_{n \to \infty} \sqrt[n]\frac{3^n n^2}{n!} = \lim\limits_{n \to \infty} \frac{3n^\frac{2}{n}}{n!^\frac{1}{n}}$$
I don't know how to calculate $\lim\limits_{n \to \infty} n!^\frac{1}{n}$. So I'm guessing maybe I am going at it in the wrong way.
 A: Checking convergence without applying a specific convergence test:
$$\sum_{n=1}^\infty \frac{3^n n^2}{n!}\leqslant\sum_{n=1}^\infty \frac{3^n\cdot3^n}{n!}=\sum_{n=1}^\infty \frac{9^n}{n!}=e^9-1<\infty$$
Also one can show with Stirlings formula that the limit $\lim\limits_{n \to \infty} n!^\frac{1}{n}$ does not exist.
A: I would suggest using the ratio test: If $\lim_{n\to\infty}\left |\frac{a_{n+1}}{a_n}\right | < 1$ we have an absolutely convergent (and in this case also convergent) series:
$$
\frac{\frac{3^{n+1}(n+1)^2}{(n+1)!}}{\frac{3^nn^2}{n!}} = \frac{3(n+1)^2}{n^2(n+1)} = \frac{3n+3}{n^2} \overset{n\to\infty}{\longrightarrow} 0
$$
Edit: How to calculate the root-check:
\begin{align}
\lim_{n\to\infty}\frac{3n^\frac{2}{n}}{n!^\frac 1 n} &= 3\lim_{n\to\infty}\sqrt[n]{\frac{n^2}{n!}} =  3\lim_{n\to\infty}\sqrt[n]{\frac{n}{(n-1)!}}\leq 3\lim_{n\to\infty}\sqrt[n]{\frac{n}{n(n-3)!}} \\
&= 3\lim_{n\to\infty}\sqrt[n]{\frac{1}{(n-3)!}} < 1
\end{align}
The inequality is correct because $(n-1)! > n(n-3)! \Leftrightarrow (n-1)(n-2) > n$ is true for $n\geq 4$. As mentioned in above comments $\sqrt[n]{n!}\to\infty$ so the term in the limit $\to 0$, it doesn't matter that a finite amount (three) of factors $n,n-1,n-2$ are missing.
We don't need to specify the exact limit, we only need to know if it is $<1$, which is shown above.
