Calculate $\lim_{n\to\infty}\binom{2n}{n}2^{-n}$ I would like to show that:
$$\lim_{n\to\infty}\binom{2n}{n}2^{-n} = \infty$$
I have gotten as far as:
$$
\binom{2n}{n}={(2n)!\over (n!)^2}=\left({n\over1}+1\right)\left({n\over2}+1\right)(\dots)\left({n\over n}+1\right)\ge2^n
$$
But the $2^{⁻n}$ factor defeats that attempt, any suggestion on how to continue would be most appreciated.
 A: Hint Keep the first bracket $({n\over1}+1)$ unchanged:
$$\binom{2n}{n}={(2n)!\over (n!)^2}=({n\over1}+1)({n\over2}+1)(\dots)({n\over n}+1)\ge2^{n-1}(n+1)$$
A: Using Stirling's approximation, we see that
$$n! \sim \frac{n^n}{e^n} \sqrt{2\pi n}$$
Thus, we have
$${2n \choose n} = \frac{(2n)!}{(n!)^2} \sim \frac{(2n)^{2n} \sqrt{4\pi n} / e^{2n}}{(n^n\sqrt{2\pi n} / e^n)^2} = \frac{4^n (n^2)^n \sqrt{2} \sqrt{2\pi n}}{n^{2n} (2\pi n)} = \frac{4^n \sqrt{\pi}}{\sqrt{n}}$$
which tends to infinity as desired.
A: Look at row $2n$ in the Pascal triangle. The sum of all terms is $2^{2n}= 4^n$. Moreover, the central binomial coefficient is the largest number in that row and so $4^n \le (2n+1){{2n} \choose n}$.
Hence
$$
{{2n} \choose n} \ge \frac{4^n}{2n+1}
$$
and so
$$
{{2n} \choose n}2^{-n} \ge \frac{2^n}{2n+1} \to \infty
$$
A: A related problem.
Hint:
You can use the result

If $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}=a$ and $|a|<1$, then $\lim_{n\to \infty}{a_n}=0 $ and the $\lim_{n\to \infty}{a_n}=\infty$ if $|a|>1$. $

A: We can use the following facts to get a lower bound on $\binom{2n}{n}$.


*

*$\sum_{k=0}^{2n}\binom{2n}{k}$=$\left ( \binom{2n}{0} + \binom{2n}{2n} \right )+ \sum_{k=1}^{2n-1} \binom{2n}{k}$ = $2^{2n}$

*The maximum value of $\binom{2n}{k}$ is attained at $\binom{2n}{n}$


Thus:
$$\binom{2n}{n} \geq \frac{2^{2n}}{2n}$$
and so:
$$\binom{2n}{n}2^{-n} \geq \frac{2^n}{2n}$$
Using L'Hospital's rule to take the limit of the RHS we get $\lim_{n\to \infty}2^{n-1}\ln2 = \infty$.
