# 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.

• The limit $\lim\limits_{n \to \infty} n!^\frac{1}{n}$ diverges to infinity. See Stirling. Jun 2, 2021 at 14:47
• Using the ratio test is simpler. Jun 2, 2021 at 14:48
• There is a result that $\lim\limits_{n\to\infty}(a_n)^{\frac{1}{n}} = \lim\limits_{n\to \infty} \frac{a_{n+1}}{a_n}$, provided the latter limit exists (assuming $a_n$ to be positive). Can you use it? Jun 2, 2021 at 14:49
• If everything isn't raised to the $n$ generally you don't want to use the root test. However, if you really wanted to, you could use Stirling's approximation en.wikipedia.org/wiki/Stirling%27s_approximation to convert $n!$ into something you can take the root of easier
– Alan
Jun 2, 2021 at 14:51
• Also $\sum_{n=1}^\infty \frac{z^n n^2}{n!} = e^z z (z + 1)$
– lhf
Jun 2, 2021 at 15:00

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.

• Best answer! +1. Jun 2, 2021 at 14:59

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.

• That last limit is not $1$ Jun 2, 2021 at 14:49
• Now it is :D The "!" was too much xd Jun 2, 2021 at 14:49