Finding the limit inf./inf. form

Question

Find the limit of the sequence defined by

Lim $_{n \to \infty} \frac{(2n)!}{(2^n\cdot n!)^2}$

How to start?

Is there another way of showing this sequence is convergent without finding the limit?

Answer 1

The asymptotic of the central binomial coefficient is given by $$\dbinom{2n}n \sim \dfrac{4^n}{\sqrt{(n+1/2)\pi}}$$ Hence, we have that $$\dfrac{(2n)!}{(2^n \cdot n!)^2} \sim \dfrac1{\sqrt{\pi (n+1/2)}}$$ This means $$\lim_{n \to \infty} \dfrac{(2n)!}{(2^n \cdot n!)^2} = 0$$

Answer 2

Using Stirling's approximation, $n! \sim \sqrt{2\pi n}\left ( \frac{n}{e}\right )^n$, we have

$$\lim_{n\rightarrow\infty}\frac{(2n)!}{(2^n\cdot n!)^2}=\lim_{n\rightarrow\infty}\frac{2\sqrt{\pi n}\left ( \frac{2n}{e}\right )^{2n}}{\left (2^n\cdot \sqrt{2\pi n}\left ( \frac{n}{e}\right )^n\right )^2}$$

Some cancelling will leave you with the answer.

Answer 3

To show the sequence is convergent without finding the limit:

• if $a_n = \dfrac{(2n)!}{(2^n\cdot n!)^2}$

• then $a_0=1$

• and $\dfrac{a_n}{a_{n-1}} = \dfrac{(2n)(2n-1)}{2^2n^2} = 1- \dfrac{1}{2n}$ which is between $0$ and $1$ for $n \ge 1$

• so $a_n$ represents a decreasing positive sequence and is therefore bounded below by $0$ and thus converges to a non-negative value

Answer 4

$$\dfrac{(2n)!}{(2^n\cdot n!)^2}=\dfrac{\prod_{r=1}^n(2r)}{2^n\cdot n!}\cdot\dfrac{\prod_{u=1}^n(2u-1)}{2^n\cdot n!}=\prod_{r=1}^n\dfrac{2r-1}{2r}$$

which is $\le\dfrac1{\sqrt{3n+1}}$ (Proof)

See also :

limit of $\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$ using geometric mean

Convergent Series 2n-1/2n

The lower bound of $\prod_{r=1}^n\dfrac{2r-1}{2r}$