Finding $\lim_{n\to\infty}\binom{2n}{n}$ I found a limit problem which is
$\lim_{n\to\infty}{2n \choose n}$
as we know that $\sum_{i=0}^{n}{n \choose i}^2={2n \choose n}$
now we could transform the given problem as $\lim_{n\to\infty} \sum_{i=0}^{n}{n \choose i}^2$
and as $\sum_{i=0}^{n}{n \choose i}^2$ is a polynomial in $n$,
therefore, $\lim_{n\to\infty}\sum_{i=0}^{n}{n \choose i}^2=\infty$
Please tell me if my solution is correct or not,and if it is incorrect then please provide me a solution.
Thanks
 A: $\binom{2n}{n}$ is called the central Binomial coefficient. Wikipedia says that the Wallis product can be written in the form of an asymptotic for the central Binomial coefficient. That is, $$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$$   

This supposedly has a short proof:
$$
{2n \choose n} = 
 \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx < 
 \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} e^{-nx^2} dx <
 \frac{4^n}{\pi} \int_{-\infty}^{\infty} e^{-nx^2} dx =
 \frac{4^n}{\sqrt{\pi n}}.
$$
In the first step, the formula for $\int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx$ can be proved by using the Beta function.  
The third step is clear, and the last step is the well-known Gaussian integral.
So we need only justify the the second step.
There we need the inequality $\cos x \leq e^{-x^2/2}$, that is,
$$
\log \cos x + \frac{x^2}{2} \leq 0,
$$
for $\left|x\right| < \pi/2$, with equality only at $x=0$.
This is true because $\log \cos x +\frac12 x^2$ is an even function that vanishes at $x=0$ and whose second derivative ($-\tan^2 x$) is negative for all nonzero $x \in (-\pi/2, \pi/2)$. Hope it helps.
A: Alternatively,
$$
f(n)= {2n\choose n} \implies \frac{f(n+1)}{f(n)}\ =\ \frac{2\:(n+1)\:(2n+1)}{\ (n+1)\ (n+1)}\ge2, \quad n \ge0,
$$ giving
$$
{2n\choose n} \ge 2^n.
$$
A: Perhaps a more elegant approach: $n^2=\displaystyle{n \choose 1}^2\le\sum_{i=0}^{n}{n \choose i}^2$ and $\displaystyle\lim_{n\to +\infty}n^2=+\infty$.
A: $$\binom{2n}{n} \ge \binom{2n}{1} = 2n \to \infty$$
A: $S=\binom{2n}{n}$
$\implies S=\frac{2^n(2n-1)(2n-3)...1}{n!}$
$\implies 2^n<S<\frac{2^n(2n)(2n-2)....2}{n!}$
$\implies 2^n<S<2^{2n}$
Now using sandwich we get the limit as $\infty$
A: $(2n)!/(n! . n!)$ numerator contains one term of denominator and remaining factor of numerator is greater than other term in denominator. And it keeps increasing. So the whole thing tends to Infinity.
