Harder limit

[ivory adder]

$ \displaystyle \lim_{n \to \infty}\frac{1}{4^n}\begin{pmatrix}
2n \
n
\end{pmatrix}$

How do I solve this?

[tough totem]

danilojonic

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[ivory adder]

$\begin{pmatrix}
2n \
n
\end{pmatrix} = \frac{2n!}{n!^2}$

[tough totem]

danilojonic

[ivory adder]

I think the form I got to is $0 \cdot 0$

[tough totem]

danilojonic

[ivory adder]

But I dont see the implementation of Lhopital for example

[slate hull]

...if it's 0 * 0, that's not an indeterminate form, it's... just 0.

But also I don't think that is what you got.

[ivory adder]

yes I think that too

[ivory adder]

yeah I know, what I wanted to say is that I dont think I got something correct

[slate hull]

What do you mean, you "don't see the implementation of L'Hopital" then?

[ivory adder]

cuz I first made a mistake and got infinity * 0 in original comment. then I fixed it but forgot to remove lhopital

[slate hull]

Something to note is that 4^n = 2^(2n).

Also, (2n)!/(4^n * (n!)^2) goes to inf/inf, so L'Hopital's rule is kind of valid here.

Though you'd need to use the gamma function to have a differentiable form of the factorial function.

But wait, that's not differentiable I don't think.

Or maybe it is, but not in the way I thought.

[sonic ibex]

You can use Stirling's approximation.

[sonic ibex]

Well, f(x) = 4^(-x) C(2x, x) is differentiable, but its derivative won't look very nice - a lot of digammas.

[slate hull]

Okay, but we're talking about using L'Hopital's rule, so we're not talking about differentiating that, we're talking about differentiating (2n)! and 4^n * (n!)^2.

[ivory adder]

I never differentiated a factorial so I have absolutely no clue how that would go

[sonic ibex]

I don't think that's really feasible to use here.

[slate hull]

Yeah, I was just saying it does technically apply, not that it was a good idea.

[sonic ibex]

I mean, if not that, we can at least prove that a(n + 1)/a(n) is less than 1 (from some n, at least).

That is pretty easy.

[polar nest]

First off I would show that the limit of 2n!/((n!)^2) is 0, which is what OP most likely showed.

we can replace n! by x, we have 2x/(x^2) and the limit is clearly 0. This is enough. Or, just divide, we have 2/n! which is clearly 0.

1/4^n has a trivial limit of 0.

The limit product rule then applies and the limit does equal 0.

[sonic ibex]

(2x)!/(x!)^2 doesn't go to 0 for x -> +∞, though. It behaves like ||4^x/√(πx).||

[polar nest]

Ah yes, my mistake, I just blindly looked at the comments

${2n \choose n} = \frac{(2n)!}{n!^2} \neq \frac{2n!}{n!^2}$

[tough totem]

EmeX

[sonic ibex]

Ah, I see.