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I am asked to determine if a series converges or not: $\displaystyle\sum\limits_{n=1}^{\infty} (2^n)n!/(n^n)$

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

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    Are you allowed to use Stirling's Approximation? It claims that for large $n$, $n! \sim \sqrt{2\pi n} e^{-n} n^n$. – 2017-02-24
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    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. – 2017-02-24
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    I would take a look in your text and see if they mention Stirling's Approximation in the relevant section. I cannot recall any other elementary ways to deal with the $n$th root of a factorial. – 2017-02-24
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    Thank you for your help. The Stirling's Approximation is not mentioned. – 2017-02-24
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    Stirling's approximation is overkill here. – 2017-02-24
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    http://math.stackexchange.com/q/28476/ – 2017-02-24
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    @MarwanNour In fact, if you have half an hour, you may sit down and write the Taylor expansion for $\ln(1+t)$ and $\ln(1-t)$ and calculate their difference $\frac12 \ln\frac{1+t}{1-t}$. Then spend 15 minutes to [read a proof of this approximation](http://www.sosmath.com/calculus/sequence/stirling/stirling.html). Since then, you can save *hours* for limits involving factorials. – 2017-02-24
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    @GNUSupporter I will look into that. Thanks again – 2017-02-24
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    If you look at [frequent questions tagged limit+factorial](http://math.stackexchange.com/questions/tagged/limits+factorial?sort=frequent), you can find [$\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite](http://math.stackexchange.com/q/136626) and [Calculating the limit $\lim((n!)^{1/n})$](http://math.stackexchange.com/q/706461). (And also some of the questions [linked there](http://math.stackexchange.com/questions/linked/136626) might be of interest.) – 2017-02-25
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    And [using Approach0](https://approach0.xyz/search/?q=%24%5Csum%5Climits_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%20%5Cfrac%7B2%5Enn!%7D%7Bn%5En%7D%24&p=1) you can also find some posts about the series mentioned in your question, for example, [here](http://math.stackexchange.com/questions/1931236/convergent-or-divergent-sum-k-1-infty-frac2k-cdot-kkk) or [here](http://math.stackexchange.com/questions/751275/simple-series-convergence-divergence). – 2017-02-25
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    Basically the situation with this post is that it contains three different questions. The title suggest that it is about $\lim\limits_{n\to\infty} \sqrt[n]{n!}$. However, in the body the OP asks about the limit $\lim\limits_{n\to\infty}\frac{\sqrt[n]{n!}}n$. And they ask about this limit because they are interested in $\sum\limits_{n=1}^{\infty} \frac{2^nn!}{n^n}$. Whether this question is considered as a question about limit or about series, it seems that there are already duplicate questions where it was answered. – 2017-02-25
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    No need to yell when titling your question. – 2017-02-25

3 Answers 3

9

Note that we do not need to actually evaluate the limit, we just need to find an upper bound.

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 $0.997932$.

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    +1 for the upper bound to avoid limits involving $e$ and log. At the 2nd step, it's $n! \le (6k)!$. – 2017-02-25
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    @Dr.MV - I didn't presume anything. Convergence isn't necessary, just that the $\limsup$ is less than 1 (and $\liminf$ isn't negative). – 2017-02-26
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    @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. – 2017-02-26
51

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.

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    Short. Concise. Love it. – 2017-02-25
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    @gnusupporter Thank you. Much appreciative! -Mark – 2017-02-25
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    Upvoted for this brilliant proof, but please stop using citations to show emphasis, for something that isn't quoted text. – 2017-02-25
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    @federicopoloni Thank you, Much appreciative. I don't know what you mean regarding citations to show emphasis. – 2017-02-25
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    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. – 2017-02-25
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    @FedericoPoloni Even site admin use `>` to highlight a sentence b/c it looks great. For example, see [Tag management 2017](http://meta.math.stackexchange.com/a/25910/290189) on Meta Math.SE – 2017-02-25
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    @GNUSupporter "Even the admins do it" does not mean it's correct. We have had a similar discussion at academia.se: http://meta.academia.stackexchange.com/questions/2200/adding-emphasis-with-blockquotes – 2017-02-25
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    @FedericoPoloni I see your point. You may start a post at meta math.se to raise the awareness of site admin. That's more effective than saying this hundreds of times in a comment. – 2017-02-25
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    @GNUSupporter Done: http://meta.math.stackexchange.com/questions/25939/stop-misuse-of-markdown-blockquotes-for-emphasis – 2017-02-25
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    @FedericoPoloni upvoted – 2017-02-25
30

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.

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

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    @MarwanNour You're welcome. – 2017-02-24
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    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 $$\frac{2^n n!}{n^n}\sim\frac{2^n\sqrt{2\pi n} e^{-n} n^n}{n^n}$$ – 2017-02-26
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    @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*. – 2017-02-26
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    Perfect. (Any idea why two answers on this page are so massively upvoted?) – 2017-02-26
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    @Did For the other one, yes: his argument using Riemann sum is *so* beautiful and he makes use of elementary tools. For mine, I have no idea. I just type the second part for fun. – 2017-02-26