- #1

Battlemage!

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## Homework Statement

(This isn't actually course work for me but the textbook is used in course work at my school)

Find the limit:

[tex] \lim_{x\rightarrow ∞}\frac{(n!)^2}{(2n)!}[/tex]

## Homework Equations

None other than what a factorial and limit are. 4! = (1)(2)(3)(4), for example.

## The Attempt at a Solution

In examining (2n)!, it looks like:

[tex](2n)! = (1)(2)(3)...(n)(n+1)...(2n-3)(2n-2)(2n-1)(2n)[/tex]

which is

[tex](n!)(n+1)...(2n-3)(2n-2)(2n-1)(2n)[/tex]

So I can cancel one n! term, leaving:

[tex]\frac{n!}{(n+1)(n+2)...(2n-2)(2n-1)(2n)}[/tex]

Then expanding n! a little bit gives:

[tex]\frac{(1)(2)...(n-2)(n-1)n}{(n+1)...(2n-2)(2n-1)(2n)}[/tex]

And it's here I notice that n+1, the first term in what's left of my denominator, is larger than 1, the first term in my numerator, for all n>0, n∈ℕ. And of course, n+2>2 for all n>0, and so on, all the way to the very end, when 2n>n for all n>0.

By this reasoning I argue that if n approaches infinity, the denominator will approach faster, and so the limit is zero.

I know zero is the correct answer, but would the denominator simply being larger be a good enough reason to declare that the limit is zero?

If not, here's my "Phase Two" plan.

If not, here's my "Phase Two" plan.

If I factor out 2

^{n}from the numerator I'd have (

*I think*!):

[tex]\frac{(1)(2)...(n-2)(n-1)n}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})n(2^n)}[/tex]

Then I divide by n,

[tex]\frac{(1)(2)...(n-2)(n-1)}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})(2^n)}[/tex]

And now I have a high degree polynomial divided by a high degree polynomial TIMES an exponential function, and as was shown in another thread where I had questions regarding that specific topic, the exponential wins by a mile. So in this case, the limit has to be zero, because the denominator approaches infinity WAY faster.

*I also just noticed that factoring out the 2

^{n}will also take out every single term in the numerator: 2n-4 = 2(n-2), so goodbye n-2 term; 2n-6 = 2(n-3), so goodbye n-3 term, and it should go that way all the way down. So I'd have 1 in the numerator and (n-(1/2))2

^{n}in the denominator, meaning this race to infinity is shamefully one sided (assuming my factoring of 2

^{n}isn't done incorrectly). Yes, no, maybe so? Any thoughts on a better way to do this?Thanks!