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 $(\sqrt[n]{n!})$?


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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    $\begingroup$ @MarwanNour You're welcome. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 24 '17 at 22:33
  • $\begingroup$ 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 $$\frac{2^n n!}{n^n}\sim\frac{2^n\sqrt{2\pi n} e^{-n} n^n}{n^n}$$ $\endgroup$ – Did Feb 26 '17 at 7:35
  • $\begingroup$ @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. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 26 '17 at 8:56
  • $\begingroup$ Perfect. (Any idea why two answers on this page are so massively upvoted?) $\endgroup$ – Did Feb 26 '17 at 9:36
  • $\begingroup$ @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. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 26 '17 at 14:36

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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  • $\begingroup$ Short. Concise. Love it. $\endgroup$ – John Smith Feb 25 '17 at 0:12
  • $\begingroup$ @gnusupporter Thank you. Much appreciative! -Mark $\endgroup$ – Mark Viola Feb 25 '17 at 0:34
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    $\begingroup$ Upvoted for this brilliant proof, but please stop using citations to show emphasis, for something that isn't quoted text. $\endgroup$ – Federico Poloni Feb 25 '17 at 12:37
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    $\begingroup$ @federicopoloni Thank you, Much appreciative. I don't know what you mean regarding citations to show emphasis. $\endgroup$ – Mark Viola Feb 25 '17 at 13:30
  • $\begingroup$ 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. $\endgroup$ – Federico Poloni Feb 25 '17 at 14:25

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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  • $\begingroup$ +1 for the upper bound to avoid limits involving $e$ and log. At the 2nd step, it's $n! \le (6k)!$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 25 '17 at 10:40
  • $\begingroup$ @Dr.MV - I didn't presume anything. Convergence isn't necessary, just that the $\limsup$ is less than 1 (and $\liminf$ isn't negative). $\endgroup$ – Glen O Feb 26 '17 at 6:37
  • $\begingroup$ @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. $\endgroup$ – Mark Viola Feb 26 '17 at 6:47

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