$ \lim_{n \to \infty} \binom{2n}{n}$

Question

I've tried to solve the limit $$ \lim_{n \to \infty} \binom{2n}{n}$$ but I'm not sure.

$ \lim_{n \to \infty} \binom{2n}{n} = \lim_{n \to \infty} \frac{(2n)!}{n! n!} $

Now I have put $\frac{(2n)!}{n! n!} \sim \frac{(2n)^{2n}}{n^n n^n } =\frac{(2n)^{2n}}{n^{2n} }= (\frac{2n}{n})^{2n}=2^{2n} \rightarrow +\infty$

Is it right?

Looking at Pascal's triangle, it is clear that it more than doubles each time, and is clearly increasing along the middle. — Simply Beautiful Art, Dec 14 '19 at 22:38

score 4 · Answer 1 · answered Dec 14 '19 at 22:35

4

This is overcomplicated. Let $M>0$. Then for $n\geq M$, $\binom{2n}{n}\geq n\geq M$ so $\lim_{n\to \infty}\binom{2n}{n}=+\infty$.

answered Dec 14 '19 at 22:35

Adam Martens

2,539

score 1 · Answer 2 · answered Dec 14 '19 at 22:39

1

Hint

$$\binom{2n}{n}= \frac{(2n)!}{n! n!} = \frac{(2n-1)2n}{n^2} \binom{2(n-1)}{n-1}\geq 2 \binom{2(n-1)}{n-1} $$

Show that this gives $$\binom{2n}{n} \geq 2^n$$

answered Dec 14 '19 at 22:39

N. S.

132,525

score 0 · Accepted Answer · answered Dec 14 '19 at 23:22

0

Defining a very rough but easy lower bound: $$\binom{2n}n=\frac{(n+1)(n+2)(n+3)\dotsb (n+n)}{n!}=(n+1)\Bigl(\frac n2+1\Bigr)\Bigl(\frac n3+1\Bigr)\dotsb \Bigl(\frac nn+1\Bigr)>n+1,$$ so it tends to $+\infty$.

answered Dec 14 '19 at 23:22

Bernard

175,478

$ \lim_{n \to \infty} \binom{2n}{n}$

3 Answers3