prove that $\lim_{n\to\infty} \sqrt[n]{\binom{2n}{n}}=4$

Question

prove that $\lim_{n\to\infty} \sqrt[n]{\binom{2n}{n}}=4$

I proved this using induction and the squeeze theorem I didn't found any one that solve it this way so i wanted to share my solution.

Answer 1

Lets show that $\frac{4^{n-1}}{n}<\binom{2n}{n}<4^n$,with induction

First lets simplify $$\binom{2n}{n}=\frac{n!(n+1)...(2n)}{n!(2n-n)!}=\frac{(n+1)(n+2)...(2n)}{n!}$$ Basis:Show that the statement hold for $n=1$ ,$1<2<4$ we got true statement
Lets assume the assumption holds for $n=k$ and lets prove its hold for $n=k+1$ $\frac{4^{k-1+1}}{k+1}<\binom{2k+2}{k+1}<4^{k+1}$. First lets solve this for the first equation we have $\frac{4^{k-1+1}}{k+1}<\binom{2k+2}{k+1}=\frac{(k+2)(k+3)...2k(2k+1)2(k+1)}{k!(k+1)}$ now lets multipy by $\frac{k+1}{4}$ and simplify the right side a little and we will get $4^{k-1}<\frac{(k+1)(k+2)(k+3)...(2k-1)(2k+1)}{(k-1)!}$ lets show that $(k+1)(k+2)(k+3)...(2k-1)(2k+1)>(k+1)(k+2)...2k$ we will eliminate the same terms from both sides and get $2k+1>2k$ which is true we have $$\frac{4^{k-1}}{k}<\frac{(k+1)(k+2)...2k}{k!}<\frac{(k+1)(k+2)...(2k-1)(2k+1)}{k!}=\binom{2k+2}{k+1}$$ as we wanted to show
Now lets solve for the second equation: $\frac{(k+2)(k+3)...2k(2k+1)2(k+1)}{k!(k+1)}<4^{k+1}$ lets simpilify the left side and multipy by $\frac{1}{4}$ Hence we got $\frac{(k+2)(k+3)...(2k-1)k(2k+1)}{k!}<4^{k}$ and because $2k+1<2(k+1)$
We got $\frac{(k+2)(k+3)...(2k-1)k(2k+1)}{k!}<\frac{(k+1)(k+2)...2k}{k!}<4^{k}$ Since both the basis and the inductive step have been performed,by mathematical induction $$\frac{4^{n-1}}{n}<\binom{2n}{n}<4^n$$ for all $1\le n$
So now we can prove that $\lim_{n\to\infty} \sqrt[n]{\binom{2n}{n}}=4$ using squeeze theorem since $\lim_{n\to\infty}\sqrt[n]\frac{4^{n-1}}{n}=\lim_{n\to\infty}\sqrt[n]{4^{n}}=4$

Answer 2

You can do it as follows you know that if $\lim_{n\to \infty} \sqrt[n]{a_n}$ exists then $\lim_{n\to \infty} \sqrt[n]{a_n} = \lim_{n\to \infty} \frac{a_{n+1}}{a_n}$. So: \begin{align} \lim_{n\to\infty} \frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}= \lim_{n\to\infty} \frac{\frac{(2n+2)!}{(n+1)!(n+1)!}}{\frac{(2n)!}{n!n!}} = \lim_{n\to\infty} \frac{\frac{(2n)!(2n+1)(2n+2)}{n!n!(n+1)^2}}{\frac{(2n)!}{n!n!}} = \lim_{n\to \infty} \frac{(2n+1)(2n+2)}{(n+1)^2} = 4 \end{align} Thus: \begin{align} \lim_{n\to\infty} \sqrt[n]{\binom{2n}{n}} = 4 \end{align}