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$$\lim_{n \to \infty}\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}$$

I was practicing exercise with a visible sum in those but how to get one here?

UfmdFkiF
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4 Answers4

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Note that we can write

$$\begin{align} \log\left(\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\right)&=\frac1n\left(\log((2n)!)-\log(n!)\right)-\log(n)\\\\ &=\frac1n\left(\sum_{k=1}^{2n}\log(k)-\sum_{k=1}^n\log(k)\right)-\log(n)\\\\ &=\frac1n\sum_{k=n+1}^{2n}\log(k)-\log(n)\\\\ &=\frac1n\sum_{k=1}^{n}\log(k+n)-\log(n)\\\\ &=\frac1n\sum_{k=1}^{n}\left(\log(k+n)-\log(n)\right)\\\\ &=\underbrace{\frac1n\sum_{k=1}^n\log(1+k/n)}_{\text{Riemann Sum for}\,\,\int_0^1 \log(1+x)\,dx=2\log(2)-1} \end{align}$$

Hence, we have

$$\lim_{n\to \infty}\left(\frac1n\sqrt[n]{\frac{(2n)!}{n!}}\right)=4e^{-1}$$

And we are done!

Mark Viola
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  • +1You deserve it, It is the actual solution to this question. – Jaideep Khare Apr 29 '17 at 19:45
  • @JaideepKhare Thank you my friend! – Mark Viola Apr 29 '17 at 19:45
  • I know that feel, when our answer is perfect but it isn't accepted. :P Don't worry sir, your answer is perfect. I don't know why some users don't accept any answer which answers them. Even you can see some of my answers have not been accepted yet, which actually answer the OP. – Jaideep Khare May 01 '17 at 07:21
  • We need $$\lim_{n\to \infty}\left(\sqrt[n]{\frac{(2n)!}{n! \times n^n}}\right)$$ not $$\lim_{n\to \infty}\left(\sqrt[n]{\frac{(2n)!}{n!}}\right)$$ Is there something I am missing? – Jaideep Khare May 02 '17 at 18:10
  • @JaideepKhare You aren't missing anything. Thank you for catching this! I've edited to correct. Much appreciative my friend. -Mark – Mark Viola May 02 '17 at 18:25
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We can also use the following theorem:

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

Let $a_{n} = (2n)!/(n!n^{n})$ and we can see that $$\frac{a_{n+1}}{a_{n}}=\frac{(2n+2)!}{(n+1)!(n+1)^{n+1}}\cdot \frac{n!n^{n}} {(2n)!}=\frac{2(2n+1)}{n+1}\cdot\frac{n^{n}}{(n+1)^{n}}\to\frac{4}{e}$$ and hence $$a_n^{1/n}=\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}}\to\frac{4}{e}$$

Paramanand Singh
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  • @Dr.MV: I fully agree with your viewpoint. But I think a person of your stature should not feel bad about it. Your answer (which I looked at just now) does deserve upvote (so there goes my +1), but I think sometimes most people don't read all the answers (sometimes I also do it). – Paramanand Singh Apr 29 '17 at 19:48
  • You're right. It's more of bewilderment. And I'm likely guilty of being inconsistent in my voting behavior too, which is even more bewildering ... LOL. – Mark Viola Apr 29 '17 at 20:50
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Use $$(2n)!=2^n \cdot n! \cdot 1 \cdot 3 \cdot 5 \ldots (2n-1)$$

Now pull $2^n$ out of the root, push $\dfrac{1}{n}$ into the root, cancel out $n!$ and then use logarithms.

You'll get :

$$A=\frac{1}{n}\sqrt[n]{\frac{(2n)!}{n!}} \implies \ln A=\ln2+\sum_{r=1}^{n}\ln\left(\frac{2r-1}{n} \right)$$

Can you proceed now (Using Integrals)?

Jaideep Khare
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  • Yea, I think I can handle it now, Thanks – UfmdFkiF Apr 27 '17 at 22:33
  • Oh, but I can't figure out why $(2n)!=2^n \cdot n! \cdot 1 \cdot 3 \cdot 5 \ldots (2n-1)$? – UfmdFkiF Apr 27 '17 at 22:45
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    @TheMeff $$(2n)!=1.2.3.4.....(2n)=(2.4.6.8...2n)(1.3.5....(2n-1))=((2.1)(2.2)(2.3)(2.4)...(2.n))(1.3.5...(2n-1))=2^{n}(1.2.3....n)(1.3.5....(2n-1))=2^{n}(n!)(1.3.5....(2n-1))$$ – Jaideep Khare Apr 27 '17 at 22:53
  • Jaideep, this post presents a way forward, but neither addresses the question as suggested by the title "Calculate limit using Riemann integral/Riemann sum," nor regards the tags "riemann-sum" and "riemann-integration." -Mark – Mark Viola Apr 29 '17 at 19:27
  • @Dr.MV But I think we can proceed and compute the last sum (which I left for OP) using Riemann Integrals.(Actually Riemann is a new term for me, I don't know anything in detail about it, but I have seen others using this term in these type of integrals, I am just a highschool student) Correct me if I am wrong. – Jaideep Khare Apr 29 '17 at 19:31
  • The last term is not in the form of a Riemann sum. Have a look at my solution. You will see a sum $\frac 1n \sum_{k=1}^n\log(1+k/n)$. In the limit as $n\to \infty$, this sum becomes the integral $\int_0^1 \log(1+x),dx=1+\log(4)$. Note that the $\frac1n$ term outside the sum is just the length of each subinterval of a uniform partition of $[0,1]$ in $n$ subintervals and $\log(1+k/n)$ is the value of $\log(1+x)$ at the right end point of the $k$'th subinterval. -Mark – Mark Viola Apr 29 '17 at 19:37
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\bracks{{1 \over n}\root[n]{\pars{2n}! \over n!}} & = \lim_{n \to \infty}\braces{{1 \over n} \bracks{\root{2\pi}\pars{2n}^{2n + 1/2}\expo{-2n} \over \root{2\pi}n^{n + 1/2}\expo{-n}}^{1/n}} \\[5mm] &= \lim_{n \to \infty}\bracks{{1 \over n}\pars{2^{2n + 1/2}\,n^{n}\expo{-n}}^{1/n}} = \expo{-1}\lim_{n \to \infty}\bracks{2^{2 + 1/\pars{2n}}} = \bbx{\ds{4 \over e}} \end{align}

Felix Marin
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