# Limiting behavior of a sequence

• Jan 22nd 2011, 09:05 PM
Glitch
Limiting behavior of a sequence
The question:

Describe the limiting behaviour of the following sequence. If the sequence converges, then state its limit.

$\displaystyle \frac{(2n)!}{(n!)^2}$

I'm not sure how to evaluate this. I tried get some intuition of what's going on, but I'm struggling. Any advice?
• Jan 22nd 2011, 09:08 PM
Chris L T521
Quote:

Originally Posted by Glitch
The question:

Describe the limiting behaviour of the following sequence. If the sequence converges, then state its limit.

$\displaystyle \frac{(2n)!}{(n!)^2}$

I'm not sure how to evaluate this. I tried get some intuition of what's going on, but I'm struggling. Any advice?

Hint: Consider using Stirling's approximation for $\displaystyle n!$.

Edit: Stirling's formula could be overkill for this. You may want to consider applying the ratio test for sequences instead.

So let $\displaystyle a_n=\dfrac{(2n)!}{(n!)^2}$. Now compute $\displaystyle \lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n} \right|\rightarrow L$. If $\displaystyle L<1$, then its convergent.

Can you proceed?
• Jan 22nd 2011, 09:21 PM
Glitch
Hmm, the ratio test seems to be much further into this chapter, which is beyond this problem set. I'll try it anyway, however It'd be interesting to know if there's another method.
• Jan 22nd 2011, 09:27 PM
Chris L T521
Quote:

Originally Posted by Glitch
Hmm, the ratio test seems to be much further into this chapter, which is beyond this problem set. I'll try it anyway, however It'd be interesting to know if there's another method.

If that's the case then, I would probably go along with my first suggestion and use Stirling's approximation: $\displaystyle n!\sim\sqrt{2\pi n}e^{-n}n^n$.

So $\displaystyle \lim\limits_{n\to\infty}\dfrac{(2n)!}{(n!)^2}\sim\ lim\limits_{n\to\infty}\dfrac{\sqrt{4\pi n}e^{-2n}(2n)^{2n}}{\left(\sqrt{2\pi n}e^{-n}n^n\right)^2}=\ldots$

Can you proceed?
• Jan 22nd 2011, 09:44 PM
Glitch
Ahh yep, that was fairly easy to work out after doing a bit of algebra. Thanks.

Odd that Stirling's approx. isn't mentioned in this text either. I think they wanted us to solve it via intuition. 0_o
• Jan 22nd 2011, 10:18 PM
FernandoRevilla
An alternative:

$\displaystyle \dfrac{(2n)!}{(n!)^2}=\dfrac{ (n+1)\cdot(n+2)\cdot \ldots\cdot (2n)}{1\cdot 2\cdot \ldots\cdot n}=\displaystyle\prod_{k=1}^{n}\left(1+\frac{n}{k} \right)\geq 2^n$

and

$\displaystyle \displaystyle\lim_{n \to{+}\infty}{2^n}=+\infty$

so, the limit of the given sequence is $\displaystyle +\infty$

Fernando Revilla
• Jan 23rd 2011, 04:54 AM
HallsofIvy
Quote:

Originally Posted by Chris L T521
Hint: Consider using Stirling's approximation for $\displaystyle n!$.

Edit: Stirling's formula could be overkill for this. You may want to consider applying the ratio test for sequences instead.

So let $\displaystyle a_n=\dfrac{(2n)!}{(n!)^2}$. Now compute $\displaystyle \lim\limits_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n} \right|\rightarrow L$. If $\displaystyle L<1$, then its convergent.

Can you proceed?

The ratio test is test for convergencd of series, not sequences.

If would write $\displaystyle \frac{(2n)!}{(n!)^2}= \frac{n!(n+1)(n+2)\cdot\cdot\cdot (2n)}{n! n!}= \frac{n+1}{1}\frac{n+2}{2}\frac{n+3}{3}\cdot\cdot\ cdot\frac{2n}{n}$
• Jan 23rd 2011, 08:12 AM
FernandoRevilla
I think Chris L T521 meant to try the ratio test to see if the limit is $\displaystyle L<1$ . In that case we would deduce that the limit of the sequence is $\displaystyle 0$ :

$\displaystyle \displaystyle\lim_{n \to{+}\infty}{\left |{\dfrac{a_{n+1}}{a_n}}\right |}=L<1 \Rightarrow \displaystyle\sum_{n=1}^{+\infty}a_n\;\textrm{conv ergent}\;\Rightarrow$

$\displaystyle l=\displaystyle\lim_{n \to{+}\infty}{a_n}=0\Rightarrow (a_n)\;\textrm{convergent}$

Fernando Revilla
• Jan 23rd 2011, 08:42 AM
Chris L T521
Quote:

Originally Posted by FernandoRevilla
I think Chris L T521 meant to try the ratio test to see if the limit is $\displaystyle L<1$ . In that case we would deduce that the limit of the sequence is $\displaystyle 0$ :

$\displaystyle \displaystyle\lim_{n \to{+}\infty}{\left |{\dfrac{a_{n+1}}{a_n}}\right |}=L<1 \Rightarrow \displaystyle\sum_{n=1}^{+\infty}a_n\;\textrm{conv ergent}\;\Rightarrow l=\displaystyle\lim_{n \to{+}\infty}{a_n}=0\Rightarrow (a_n)\;\textrm{convergent}$

Fernando Revilla

Yes, that's what I was trying to get at. Sorry that I wasn't explicit about that earlier. I was in essence using the following theorem:

Theorem: Suppose that $\displaystyle (s_n)$ is a sequence of positive terms and that the sequence of ratios $\displaystyle (s_{n+1}/s_n)$ converges to $\displaystyle L$. If $\displaystyle L<1$, then $\displaystyle \lim s_n = 0$.