Limit $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$

Question

$\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$

I could not approach it beyond these simple steps,

$\lim_{n\to \infty} \frac{1}{2n} \log(\frac{2n!}{(n!)^2})$
$=\lim_{n\to \infty} \frac{1}{2n} [\log(2n)+\cdots +\log(n+1)-\log(n)-\cdots-\log1]$
$=\lim_{n\to \infty} (\log(2n)^{1/2n}+\cdots+\log(n+1)^{1/2n}-\log(n)^{1/2n}-\cdots-\log1^{1/2n})$

Now,I understand that I have to create a sum of limit and produce an integration or use the formula $\lim_{n\to \infty} \log(1+\frac1x)^x=e$ but I cannot do it. Please help!

Answer 1

Stirling's Asympotic Approximation yields $$ \binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{1} $$ and $$ \begin{align} \lim_{n\to\infty}\frac1{2n}\log\left(\frac{4^n}{\sqrt{\pi n}}\right) &=\lim_{n\to\infty}\left(\frac{n\log(4)}{2n}-\frac{\log(\pi n)}{4n}\right)\\ &=\log(2)\tag{2} \end{align} $$

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We can also approach this in a more elementary manner: $$ \begin{align} \binom{2n}{n} &=\frac{2^n}{n!}\frac{2n(2n-1)(2n-2)(2n-3)\cdots4\cdot3\cdot2\cdot1}{2n(2n-2)(2n-4)(2n-6)\cdots8\cdot6\cdot4\cdot2}\\ &=\frac{2^n}{n!}(2n-1)(2n-3)\cdots3\cdot1\\ &=4^n\frac{(2n-1)(2n-3)\cdots3\cdot1}{2n(2n-2)\cdots4\cdot2}\tag{3} \end{align} $$ To bound $(3)$, we have $$ \begin{align} \frac{4^n}{2} &=4^n\frac{2n(2n-2)\cdots6\cdot4}{2n(2n-2)\cdots4\cdot2}\\ &\ge\color{#C00000}{4^n\frac{(2n-1)(2n-3)\cdots5\cdot3}{2n(2n-2)\cdots4\cdot2}}\\ &\ge4^n\frac{(2n-2)(2n-4)\cdots4\cdot2}{2n(2n-2)\cdots4\cdot2}\\ &=\frac{4^n}{2n}\tag{4} \end{align} $$ Combining $(3)$ and $(4)$ yields $$ \frac1{2n}\log\left(\frac{4^n}{2n}\right) \le\frac1{2n}\log\binom{2n}{n} \le\frac1{2n}\log\left(\frac{4^n}{2}\right)\tag{5} $$ that is $$ \log(2)-\frac{\log(2n)}{2n}\le\frac1{2n}\log\binom{2n}{n}\le\log(2)-\frac{\log(2)}{2n}\tag{6} $$ and by the Squeeze Theorem, we get $$ \lim_{n\to\infty}\frac1{2n}\log\binom{2n}{n}=\log(2)\tag{7} $$

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More Accuracy

Stirling's Approximation actually gives $$ \binom{2n}{n}\le\frac{4^n}{\sqrt{\pi n}}\tag{8} $$ and inequality $(4)$ combined with the AM-GM yields $$ \frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\tag{9} $$ So we have the bounds $$ \frac{4^n}{2\sqrt{n}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi n}}\tag{10} $$

Answer 2

HINT:

As $\displaystyle \binom{2n}n=\frac{(2n)!}{n! n!}=\prod_{1\le r\le n}\frac{n+r}r$

$$\ln\binom{2n}n=\sum_{1\le r\le n}\ln\left(\frac{n+r}r\right)=\sum_{1\le r\le n}\ln\left(\frac{1+\frac rn}{\frac rn}\right)$$

As $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

$$\lim_{n\to\infty}\frac1n\ln\binom{2n}n=\int_0^1\ln\left(\frac{1+x}x\right)dx$$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ For $N \gg 1$, ${{\rm d}\ln\left(N!\right) \over {\rm d}N} \approx \ln\left(N\right)$. Consider $N$ and/or $n$ as real variables !!!. $$ \color{#0000ff}{\large\lim_{n \to \infty}{1 \over 2n}\,\left\{\ln\left(\left[2n\right]!\right) - 2\ln\left(n!\right)\right\}} = \lim_{n \to \infty}{2\ln\left(2n\right) -2\ln\left(n\right)\over 2} = \color{#0000ff}{\large\ln\left(2\right)} $$

Otherwise, \begin{align} &\lim_{x \to \infty}{\ln\Gamma\pars{2x + 1} - 2\ln\Gamma\pars{x + 1} \over 2x} = \lim_{x \to \infty}{2\Psi\pars{2x + 1} - 2\Psi\pars{x + 1} \over 2} \\[3mm]&= \lim_{x \to \infty}\bracks{\Psi\pars{2x} + {1 \over 2x} - \Psi\pars{x} + {1 \over x}} = \lim_{x \to \infty}\bracks{\Psi\pars{2x} - \Psi\pars{x}} \end{align} Since $\Psi\pars{z} \sim \ln\pars{z}$ when $\verts{z} \gg 1$, we'll have $\Psi\pars{2x} - \Psi\pars{x} \sim \ln\pars{2x} - \ln\pars{x} = \ln\pars{2}$.

$\Gamma$ and $\Psi$ are the ${\it Gamma}$ and ${\it Digamma}$ functions, respectively: $\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}$.

${\large\tt ADENDUM:}$ Since $\ds{\totald{\ln\pars{x}}{x} = {1 \over x}}$, the ${\large\tt\ln}$ function varies slowly when $x \gg 1$. Then, for large $N$: $$ \left.\totald{\ln\pars{x}}{x}\right\vert_{x\ =\ N} \approx {\ln\pars{\bracks{N + 1}!} - \ln\pars{N!} \over \pars{N + 1} - N} = \ln\pars{N + 1} \approx \ln\pars{N}\,,\qquad N \gg 1 $$

Answer 4

As $$ \log\binom{2n}{n}=\log\left(\frac{(2n)!}{(n!)^2}\right)=\log((2n)!)-2\log(n!) $$ using Stirling's approximation $\log(k!)\sim k\log k -k$ $$ \log((2n)!)-2\log(n!)\sim 2n\log(2n)-2n\log n=2n\log 2 $$ So $$ \lim_{n\to\infty}\frac{1}{2n}\log\binom{2n}{n}= \lim_{n\to\infty}\frac{1}{2n}2n\log 2=\log 2 $$

Answer 5

With Stolz-Cesaro (${\rm L}=\log$, obviously): $$\lim_{n\to\infty}{{\rm L}(2n)+\cdots +{\rm L}(n+1)-{\rm L}(n)-\cdots-{\rm L}1\over 2n}=$$

$$\lim_{n\to\infty}{({\rm L}(2n+2)+\cdots +{\rm L}(n+2)-{\rm L}(n+1)-\cdots-{\rm L}1)-({\rm L}(2n)+\cdots +{\rm L}(n+1)-{\rm L}(n)-\cdots-{\rm L}1)\over 2(n+1)-2n}$$ $$=\lim_{n\to\infty}{{\rm L}(2n+2)+{\rm L}(2n+1)-2{\rm L}(n+1)\over 2}= {1\over2}\lim_{n\to\infty}{\rm L}\left({(2n+2)(2n+1)\over(n+1)^2}\right)= {1\over2}{\rm L}(4)={\rm L}(2).$$

Answer 6

$Let \lim_{n\to \infty} \frac{1}{2n} log{2n\choose n}=L$
This limit can be solved by Squeeze Theorem, by the following inequality
$\lim_{n\to \infty} \frac{1}{2n} log{\frac{4^n}{n+1}}\le L\le\lim_{n\to \infty} \frac{1}{2n} log{4^n}$
Now,upon a little manipulation,left limit tends to $log2$ and right limit tends to $log2$, so $L=log2$ Hence solved.

Also, the proof of inequality can be obtained by induction hypothesis.
Sorry, could not provide full solution but it is also and idea!