A calculus problem by Yuri Lombardo

Solution as suggested by @Yuri Lombardo :

$\begin{aligned} L & = \lim_{n \to \infty} \sqrt [n] {2n \choose n} \\ & = \lim_{n \to \infty} \sqrt[n]{\frac {(2n)!}{(n!)^2}} & \small {\color{#3D99F6}\text{By Stirling's formula: }n! \sim \sqrt{2\pi n}\left(\frac ne \right)^n} \\ & = \lim_{n \to \infty} \sqrt[n]{\frac {\sqrt{4\pi n}\left(\frac {2n}e \right)^{2n}}{2\pi n \left(\frac ne \right)^{2n}}} \\ & = \lim_{n \to \infty} \sqrt[n]{\frac {2^{2n}}{\sqrt {\pi n}}} \\ & = \lim_{n \to \infty} 4(\pi n)^{-\frac 1{2n}} \\ & = \lim_{n \to \infty} 4 \exp \left(\ln (\pi n)^{-\frac 1{2n}} \right) & \small {\color{#3D99F6} \exp(x) = e^x} \\ & = \lim_{n \to \infty} 4 \exp \left(-\frac {\ln (\pi n)}{2n} \right) & \small {\color{#3D99F6}\text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.}} \\ & = \lim_{n \to \infty} 4 \exp \left(-\frac {\frac 1n}{2} \right) & \small {\color{#3D99F6}\text{Differentiate up and down w.r.t. }x} \\ & = 4e^0 = \boxed{4} \end{aligned}$

Reference: L'Hôpital's rule

---

Alternative solution

$\begin{aligned} L & = \lim_{n \to \infty} \sqrt [n] {2n \choose n} \\ & = \lim_{n \to \infty} \sqrt[n]{\frac {(2n)!}{(n!)^2}} \\ & = \lim_{n \to \infty} \exp \left(\frac 1n \ln \left(\frac {(2n)!}{(n!)^2} \right) \right) \\ & = \lim_{n \to \infty} \exp \left(\frac 1n \left(\sum_{k=1}^{2n} \ln k - 2 \sum_{k=1}^{n} \ln k \right) \right) \\ & = \lim_{n \to \infty} \exp \left(\frac 1n \left(\sum_{k=n+1}^{2n} \ln k - \sum_{k=1}^{n} \ln k \right) \right) \\ & = \lim_{n \to \infty} \exp \left(\frac 1n \left(\sum_{k=n+1}^{2n} (\ln k - \ln n) - \sum_{k=1}^{n} (\ln k - \ln n) \right) \right) \\ & = \lim_{n \to \infty} \exp \left(\frac 1n \left(\sum_{k=n+1}^{2n} \ln \frac kn - \sum_{k=1}^{n} \ln \frac kn \right) \right) & \small {\color{#3D99F6}\text{By Riemann sums}} \\ & = \lim_{n \to \infty} \exp \left(\int_1^2 \ln x \ dx - \int_0^1 \ln x \ dx \right) \\ & = \lim_{n \to \infty} \exp \left(\bigg[x \ln x - 1\bigg]_1^2 - \bigg[x \ln x - 1\bigg]_0^1 \right) \\ & = \lim_{n \to \infty} \exp \left(2 \ln 2 - \ln 1 - \ln 1 + {\color{#3D99F6} \lim_{x \to 0}x\ln x} \right) & \small {\color{#3D99F6}\text{By L'Hôpital's rule}} \\ & = \lim_{n \to \infty} \exp \left(2 \ln 2 - 0 - 0 + {\color{#3D99F6}0} \right) \\ & = e^{\ln 4} = \boxed{4} \end{aligned}$

Reference: Riemann sums

Let $a_n = { 2n \choose n}$ .
We have the recurrence relation $a_n = \frac{ 2n \times (2n-1) } { n \times n} \times a_{n-1} = (4 - \frac{2}{n} ) \times a_n$ . Because each time we are "multiplying by close to 4", it makes sense that the final limit will be 4. Let's rigorize this idea.

Let $b_n = \log a_n$ . The expression in the problem is equivalent to

$\lim_{n\rightarrow \infty} \sqrt[n]{ { 2n \choose n } } = e ^ { \lim \frac{1}{n} b_n } .$

From the recurrence relation, $b_n = \ln \left( 4 - \frac{ 2}{n} \right) + b_{n-1}$ .
Recall that (you should know / be able to prove that) if $\lim c_n = C$ , then $\lim \frac{1}{n} \sum_{i=1}^n c_i = C$ .
Hence, since $\lim \ln \left ( 4 - \frac{2}{n} \right) = \ln 4$ , it follows that $\lim \frac{1}{n} b_n = \ln 4$ .

Thus, the limit in the question is equal to $e ^ { \ln 4 } = 4$ .

Hint: use Stirling's approximation for factorials.

for $n \rightarrow + \infty$

$n! = \sqrt{2\pi n}(\frac{n}{e})^{n}$