How to evaluate $\lim_{n \rightarrow \infty}\left(\frac{(2n)!}{n!n^n} \right)^{1/n}$?

Question

Find$$\lim_{n \rightarrow \infty}\left(\frac{(2n)!}{n!n^n} \right)^{1/n}$$ is there some trick in this questions. seems it must simplify to something but I am unable to solve it.

Answer 1

$$\begin{eqnarray*}\lim_{n\to +\infty}\left(\frac{(2n)!}{n!n^n}\right)^{\frac{1}{n}}&=& \lim_{n\to +\infty}\left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right]^{\frac{1}{n}}\\&=&\exp\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\\&=&\exp\int_{0}^{1}\log(1+x)\,dx\\&=&\exp\left(\log(4)-1\right)=\color{red}{\frac{4}{e}}\end{eqnarray*}$$ by employing the continuity of the exponential/logarithm function and a Riemann sum.

Answer 2

Using Stirling's approximation, just for the sake of it and for reference (and to give an alternative approach to that of the excellent answer by lab bhattacharjee). Caveat: overly detailed.

$$ n! \operatorname*{\sim}_{n\to\infty} \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \tag{Stirling's approximation} $$ yields $$ \frac{(2n)!}{n!n^n} \operatorname*{\sim}_{n\to\infty} \frac{\sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n n^n} = \frac{\sqrt{2}\left(\frac{2}{e}\right)^{2n}}{\left(\frac{1}{e}\right)^n}= \sqrt{2}\left(\frac{4}{e}\right)^{n} $$ i.e. $$ \frac{(2n)!}{n!n^n} = \sqrt{2}\left(\frac{4}{e}\right)^{n} + o\left(\left(\frac{4}{e}\right)^{n}\right). $$

From there, $$\begin{align} \left(\frac{(2n)!}{n!n^n}\right)^{1/n} &= \exp\left(\frac{1}{n}\ln\left(\frac{(2n)!}{n!n^n}\right)\right) = \exp\left(\frac{1}{n}\ln\left(\sqrt{2}\left(\frac{4}{e}\right)^{n} + o\left(\left(\frac{4}{e}\right)^{n}\right)\right)\right) \\ &= \exp\left(\frac{1}{n}\ln\left(\sqrt{2}\left(\frac{4}{e}\right)^{n}\right)+\frac{1}{n}\ln(1+o(1)\right) \\ &= \exp\left(\frac{1}{n}\ln\left(\left(\frac{4}{e}\right)^{n}\right)+\frac{1}{n}\ln(\sqrt{2})+\frac{1}{n}\ln(1+o(1)\right) \\ &= \exp\left(\ln\left(\frac{4}{e}\right)+o(1)+o(1)\right) \\ &= \left(\frac{4}{e}\right)e^{o(1)} \\ &\xrightarrow[n\to\infty]{} \boxed{\frac{4}{e}}. \end{align}$$

Answer 3

Just use the following result:

If $a_{n} $ is a sequence of positive terms and $a_{n+1}/a_{n}\to L$ then $a_{n} ^{1/n}\to L$.

The above is easily proved by taking logs and applying Cesaro-Stolz. For the current question we take $a_{n} = (2n)!/(n!n^{n})$ and then we can see that $$\frac{a_{n+1}}{a_{n}}=\frac{(2n+1)(2n+2)}{(n+1)^{2}}\cdot\dfrac{1}{\left(1+\dfrac{1}{n}\right)^{n}}\to\frac{4}{e}$$ so the desired limit is $4/e$.

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On request of user "S.H.W" (via comment) I provide proof of the result used in this answer. Let $b_{n} =\log a_{n} $ then $b_{n} $ is well defined as $a_{n} $ is positive. Now we can apply Cesaro-Stolz on $b_{n} /n$ to get the relation $$\lim_{n\to\infty} \frac{b_{n}} {n} =\lim_{n\to\infty} \frac{b_{n+1}-b_{n}}{n+1-n}$$ provided the limit on right hand side exists finitely or infinitely. This means that $$\lim_{n\to\infty} \log(a_{n} ^{1/n})=\lim_{n\to\infty}\log(a_{n+1}/a_{n})$$ and on exponentiating (which is a continuous operation) we get $$\lim_{n\to\infty} a_{n} ^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$$ provided the limit on right hand exists finitely or infinitely. And thus the proof is completed via Cesaro-Stolz.

Most textbooks provide proofs without the use of Cesaro-Stolz and logarithms and you should have a look at that proof also because it is typical of the proofs used in the theory of limits of sequences. Roughly the idea is that if the ratio $a_{n+1}/a_{n}$ lies between $L-\epsilon $ and $L+\epsilon$ for all $n\geq m$ then $a_{n} /a_{m} $ lies between $(L-\epsilon) ^{n-m} $ and $(L+\epsilon) ^{n-m} $.

Answer 4

HINT:

$$\ln f=\lim_{n\to\infty}\dfrac1n\ln\dfrac{(2n)!}{n^n\cdot n!}$$

$$\dfrac{(2n)!}{n^n\cdot n!}=\dfrac{\prod_{r=1}^n(r+n)}{n^n}=\prod_{r=1}^n\left(1+\dfrac rn\right)$$

Now use the idea of How to represent this limit?

Answer 5

Note that $$ \left(\frac{(2n)!}{n!n^n} \right)^{1/n}=\left(\binom{2n}{n}\frac{n!}{n^n}\right)^{1/n}. $$ Now, thanks to Wallis product, we have $$ \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}. $$ On a similar line, considering Stirling approximation of $n!$, we obtain $$ \left(\frac{(2n)!}{n!n^n} \right)^{1/n}\sim \left(\frac{4^n}{\sqrt{\pi n}}\cdot \frac{\sqrt{2\pi n}(n/e)^n}{n^n}\right)^{1/n}=\frac{4}{e}(\sqrt{2})^{1/n}\to \frac{4}{e}. $$

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With Stolz-Ces$\mrm{\grave{a}}$ro Theorem:

\begin{align} &\lim_{n \to\ \infty}\ln\pars{\braces{\bracks{2n}! \over n!\,n^{n}}^{1/n}} = \lim_{n \to\ \infty} {\ln\pars{\bracks{2n}!} - \ln\pars{n!} - n\ln\pars{n} \over n} \\[5mm] = &\ \lim_{n \to\ \infty} \left\{\vphantom{\Large A}\bracks{\vphantom{\large A}\ln\pars{2} + \ln\pars{n + 1} + \ln\pars{2n + 1}} - \ln\pars{n + 1}\right. \\[2mm] &\ \left.\phantom{\,\lim_{n \to \infty}} - \bracks{\vphantom{\large A}n\ln\pars{n + 1} + \ln\pars{n + 1} - n\ln\pars{n}}\right\} \\[5mm] = &\ \lim_{n \to \infty}\left[2\ln\pars{2} + \ln\pars{n} + \ln\pars{1 + {1 \over 2n}} - n\ln\pars{n} - n\ln\pars{1 + {1 \over n}} - \ln\pars{n} - \ln\pars{1 + {1 \over n}}\right. \\[2mm] &\ \left.\phantom{\lim_{n \to \infty}}+ n\ln\pars{n}\right] \\[5mm] = &\ \ln\pars{4}\ - \ \underbrace{\lim_{n \to \infty}\bracks{n\ln\pars{1 + {1 \over n}}}} _{\ds{\to\ 1\ \mbox{as}\ n\ \to\ \infty}}\ +\ \underbrace{\lim_{n \to \infty} \bracks{\ln\pars{1 + {1 \over 2n}} - \ln\pars{1 + {1 \over n}}}} _{\ds{\to\ 0\ \mbox{as}\ n\ \to\ \infty}}\ =\ \bbx{\ln\pars{4} - 1} \end{align}

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$$ \lim_{n \to\ \infty}\bracks{\pars{2n}! \over n!\,n^{n}}^{1/n} = \exp\pars{\ln\pars{4} - 1} = \bbox[#ffe,10px,border:1px dotted navy]{\ds{4 \over \expo{}}} $$

Answer 7

The easiest way to solve the limit with high school knowledge is by realizing that the limit of the n-th root of a succession equals the limit of the ratio of two consecutive terms. EDIT: provided the limit of the ratio exists.