# Limit of $n$th root of $n$!

I am asked to determine if a series converges or not:

$$\displaystyle\sum\limits_{n=1}^{\infty} \frac{(2^n)n!}{(n^n)}$$

So I'm using the $$n$$th root test and came up with $$\lim_{n \to {\infty}}\frac{2}{n}\times(\sqrt[n]{n!})$$ I know that the limit of $$\frac{2}{n}$$ goes to $$0$$ when $$n$$ goes to infinity but what about the $$(\sqrt[n]{n!})$$?

• Are you allowed to use Stirling's Approximation? It claims that for large $n$, $n! \sim \sqrt{2\pi n} e^{-n} n^n$.
– DKS
Feb 24, 2017 at 21:47
• to be honest i don't know the stirling's approximation, I'm a college student majoring in physics and I'm taking Calculus III as a math course. Feb 24, 2017 at 21:49
• Stirling's approximation is overkill here. Feb 24, 2017 at 22:00
• If you look at frequent questions tagged limit+factorial, you can find $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite and Calculating the limit $\lim((n!)^{1/n})$. (And also some of the questions linked there might be of interest.) Feb 25, 2017 at 1:51
• And using Approach0 you can also find some posts about the series mentioned in your question, for example, here or here. Feb 25, 2017 at 1:54

I thought it might be instructive to present an approach that relies on elementary tools only. To that end, we now proceed.

Note that we can write

\begin{align} \log\left(\frac{\sqrt[n]{n!}}{n}\right)&=\frac1n\log(n!)-\log(n)\\\\ &=\frac1n\sum_{k=1}^n\log(k)-\log(n)\\\\ &=\underbrace{\frac1n\sum_{k=1}^n\log(k/n)}_{\text{Riemann Sum for}\,\,\int_0^1 \log(x)\,dx=-1}\\\\ \end{align}

Hence, we have

$$\lim_{n\to \infty}\left(\frac{2\sqrt[n]{n!}}{n}\right)=2e^{-1}$$

And we are done!

Tools Used. Straightforward arithmetic and Riemann sums.

• Short. Concise. Love it. Feb 25, 2017 at 0:12
• @gnusupporter Thank you. Much appreciative! -Mark Feb 25, 2017 at 0:34
• Upvoted for this brilliant proof, but please stop using citations to show emphasis, for something that isn't quoted text. Feb 25, 2017 at 12:37
• @federicopoloni Thank you, Much appreciative. I don't know what you mean regarding citations to show emphasis. Feb 25, 2017 at 13:30
• See those lines with yellow background, that you produced by starting a line with >? That is semantic markup for citations, in Markdown. You are supposed to use them only when you are quoting text from someone else, not to make the sentence stand out. Feb 25, 2017 at 14:25

Since OP is taking Calculus III, perhaps the ratio test from calculus II is a more suitable way.

Let $a_n = (2^n)n!/n^n$.

\begin{align} \frac{a_{n+1}}{a_n} &= \frac{(2^{n+1})(n+1)!/(n+1)^{n+1}}{(2^n)n!/n^n} \\ &= 2(n+1) \frac{n^n}{(n+1)^n} \frac1{n+1} \\ &= 2 \frac1{\left(1+\frac{1}{n}\right)^n} \\ \end{align}

$$L = \lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \lim_{n \to +\infty} 2 \frac1{\left(1+\frac{1}{n}\right)^n} = \frac2e < 1$$

So the series converges.

Alternative method by Stirling's approximation

I type this for fun and to show the power of this formula for $\sum\limits_{n=1}^{\infty} \frac{2^n n!}{n^n}$

Use the root test on $a_n = (2^n)n!/n^n$.

\begin{align} a_n =& \frac{2^n n!}{n^n} \\ \sim& \frac{2^n\sqrt{2\pi n} e^{-n} n^n}{n^n} \\ =& \sqrt{2\pi} \cdot \frac{2^n}{e^n} \cdot \sqrt{n} \end{align}

The limit $1 \le \sqrt{n}^{1/n} \le n^{1/n} \to 1$ as $n \to +\infty$ allows us to recover the ratio $2/e$ in the previous section.

$$L = \lim\limits_{n\to+\infty} a_n^{1/n} = \lim\limits_{n\to+\infty} \frac2e \sqrt{n}^{1/n} = \frac2e$$

• @MarwanNour You're welcome. Feb 24, 2017 at 22:33
• Even replacing $i$ by $n$, the formulation $$\sum\limits_{i=1}^{\infty} \frac{2^n n!}{n^n}\sim\sum\limits_{i=1}^{\infty} \frac{(2^n)\sqrt{2\pi n} e^{-n} n^n}{n^n}$$ is faulty and should read $$\frac{2^n n!}{n^n}\sim\frac{2^n\sqrt{2\pi n} e^{-n} n^n}{n^n}$$
– Did
Feb 26, 2017 at 7:35
• @Did Thanks for pointing out this logic error. I've edited my post in response to your comment. Since the convergence of the series is to be proved, this formulation is a flawed argument. Feb 26, 2017 at 8:56
Consider that, for $n>m$, $$\frac{n!}{m!}\leq n^{n-m}$$ As such, if we let $n=6k-a$, where $0\leq a\leq5$, we can observe that $$n!\leq (6k)!\leq \prod_{i=1}^6 (ik)^k=(720k^6)^k<\left (\frac{20}{6^4}\right)^k(6k)^{6k}$$ Therefore, $$\sqrt[n]{n!}<\left(\frac{20}{6^4}\right)^{(n+a)/6n}(n+a)^{1+a/n}$$ and thus $$\frac{2\sqrt[n]{n!}}{n}<2\left(\frac{20}{6^4}\right)^{1/6}\left(\frac{20}{6^4}\right)^{a/6n}(1+a/n)(n+a)^{a/n}$$ Now, as $a$ cannot be larger than 5, we can easily take the limit of each term as $n\to\infty$, to give $$\lim_{n\to\infty}\frac{2\sqrt[n]{n!}}{n}<2\left(\frac{20}{6^4}\right)^{1/6}\approx 0.997932$$ Therefore, as the limit is less than 1, it converges.
Note that the $\lim$ in the final line isn't strictly correct notation, as we have not proven that the limit exists. That said, it captures the intent, that for sufficiently large $n$, the expression will be less than $0.997932$.
• +1 for the upper bound to avoid limits involving $e$ and log. At the 2nd step, it's $n! \le (6k)!$. Feb 25, 2017 at 10:40
• @Dr.MV - I didn't presume anything. Convergence isn't necessary, just that the $\limsup$ is less than 1 (and $\liminf$ isn't negative). Feb 26, 2017 at 6:37
• @GlenO Yes, but you wrote $\lim_{n\to \infty}\frac{2\sqrt{n!}}{n}<2(20/6^4)^6$. If the limit fails to exist, then it does really make sense to write a bound. You might consider writing $\limsup$ instead of $\lim$ here. Anyway, I have up voted your answer. So, well done. Feb 26, 2017 at 6:47