Question Number 205885 by hardmath last updated on 01/Apr/24
$$\mathrm{Find}:\:\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{}\begin{pmatrix}{\mathrm{2n}}\\{\:\:\mathrm{n}}\end{pmatrix}\:=\:? \\ $$
Answered by MM42 last updated on 03/Apr/24
![(((2n)),(( n)) )=(((2n)(2n−1)(2n−2)...(n+3)(n+2)(n+1))/(n(n−1)(n−2)...3×2×1)) a=()^(1/n) (((2n)),(( n)) ) ⇒lna=(1/n)[ln(2)+ln(2+(1/(n−1)))+ln(2+(2/(n−2)))+...+ln(2+((n−3)/3))(2+((n−2)/2))(2+((n−1)/1)) =(1/n)Σ_(i=0) ^(n−1) ln(2+((i/n)/(1−(i/n)))) ⇒lim_(n→∞) a_n =∫_0 ^( 1) ln(2+(x/(1−x)))fx =∫_0 ^( 1) [ln(2−x)−ln(1−x)]dx =(x−2)ln(2−x)−(x−1)ln(1−x))]_0 ^1 =2ln2=ln4⇒lim_(n→∞) a_n =4 ✓](https://www.tinkutara.com/question/Q205982.png)
$$\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)…\left({n}+\mathrm{3}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right)}{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$${a}=\sqrt[{{n}}]{}\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\ $$$$\Rightarrow{lna}=\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{2}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)+{ln}\left(\mathrm{2}+\frac{\mathrm{2}}{{n}−\mathrm{2}}\right)+…+{ln}\left(\mathrm{2}+\frac{{n}−\mathrm{3}}{\mathrm{3}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{2}}{\mathrm{2}}\right)\left(\mathrm{2}+\frac{{n}−\mathrm{1}}{\mathrm{1}}\right)\right. \\ $$$$=\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{ln}\left(\mathrm{2}+\frac{\frac{{i}}{{n}}}{\mathrm{1}−\frac{{i}}{{n}}}\right) \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} {a}_{{n}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\mathrm{2}+\frac{{x}}{\mathrm{1}−{x}}\right){fx} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left[{ln}\left(\mathrm{2}−{x}\right)−{ln}\left(\mathrm{1}−{x}\right)\right]{dx} \\ $$$$\left.=\left.\left({x}−\mathrm{2}\right){ln}\left(\mathrm{2}−{x}\right)−\left({x}−\mathrm{1}\right){ln}\left(\mathrm{1}−{x}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}{ln}\mathrm{2}={ln}\mathrm{4}\Rightarrow{lim}_{{n}\rightarrow\infty} {a}_{{n}} =\mathrm{4}\:\checkmark \\ $$$$ \\ $$
Answered by Frix last updated on 01/Apr/24
$${n}\rightarrow\infty\:\Rightarrow\:{n}!\rightarrow\left(\frac{{n}}{\mathrm{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}}\:\Rightarrow \\ $$$$\sqrt[{{n}}]{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\rightarrow\frac{\mathrm{4}}{\:\sqrt[{{n}}]{\pi{n}}}=\mathrm{4} \\ $$