Combinatorial Limit

Another approach O f solving this is by Definite Integration as Limits of sum. But here we can't see any summation .In problems with exponents $ln$ is the key.

Let $A= \large \lim_{n \rightarrow \infty} { 2n \choose n } ^ { 1 / n }$

Taking $ln$ both sides we get $ln(A)= ln( \lim_{n \rightarrow \infty} { 2n \choose n } ^ { 1 / n } )$ $ln(A)= \lim_{n \rightarrow \infty}ln( { 2n \choose n } ^ { 1 / n } )$ $ln(A)= \lim_{n \rightarrow \infty} \frac{1}{n} ln \frac{(n+1)(n+2)...(n+n)}{1.2.3....n}$ $ln(A)= \lim_{n \rightarrow \infty} \frac{1}{n}( ln \frac{n+1}{1}+ ln \frac{n+2}{2}+ ln \frac{n+3}{3}+.....+ ln \frac{n+n}{n})$ $ln(A)= \lim_{n\rightarrow\infty}\frac{1}{n} \sum_{r=1}^{n}ln\frac{(n+r)}{r}$ $ln(A)= \int _{ 0 }^{ 1 }{ ln(\frac { 1+x }{ x } )dx }$ $ln(A)= ln(4)$ $A=4$

We know that $S(n) = {2n \choose n }$ is the largest of the $n+1$ binomial coefficients $2n \choose 0$ , $2n \choose 1$ ,...., $2n \choose 2n$ . Also, the sum of all binomial coefficients is $2^{2n} = 4^n$ So $4^n \geq S(n) \geq \frac {4^n}{2n+1}$ .

Therefore $4\geq S(n)^{\frac {1}{n}} \geq \frac {4}{{(2n+1)}^{\frac {1}{n}}}$ . The squeeze theorem now gives our result, as $\lim_{n \to \infty} (2n+1)^{\frac {1}{n}} =1$ .

We have $\dbinom{2n}{n}=\dfrac{(2n)!}{(n!)^2}$ . Subbing Stirling and simplifying gives us $\lim_{n\to\infty}\dbinom{2n}{n}^{1/n}=\lim_{n\to\infty}\dfrac{4}{\pi^{1/2n}n^{1/2n}}.$ As $n$ approaches infinity $\pi^{1/2n}$ approaches $1$ . Also $n^{1/2n}$ approaches $1$ because $n$ grows linearly while $2n$ -th root reduces exponentially. So the answer is $\fbox 4$ .