# Thread: Limit question

1. ## Limit question

Hi

Calculate the limit:

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

Where do I go wrong?

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

Thx

2. $(2n)!\ =\ 2^n\cdot n!\cdot1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$

3. Originally Posted by TheAbstractionist
$(2n)!\ =\ 2^n\cdot n!\cdot1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$

I am not really with you on why this is true.

4. Okay. Let’s take 20! for example.

$(2\times10)!$

$=\quad1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\ cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15\cd ot16\cdot17\cdot18\cdot19\cdot20$

$=\quad2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdo t16\cdot18\cdot20\cdot1\cdot3\cdot5\cdot7\cdot9\cd ot11\cdot13\cdot15\cdot17\cdot19$

$=\quad2^{10}(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7 \cdot8\cdot9\cdot10)\cdot1\cdot3\cdot5\cdot7\cdot9 \cdot11\cdot13\cdot15\cdot17\cdot19$

$=\quad2^{10}\cdot10!\cdot1\cdot3\cdot5\cdot7\cdot9 \cdot11\cdot13\cdot15\cdot17\cdot19$

Yes?