Is this acceptable reasoning for the lim n^2/(2n) as n->∞

[Battlemage!]

Homework Statement

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

Find the limit:
$$\lim_{x\rightarrow ∞}\frac{(n!)^2}{(2n)!}$$

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:

$$(2n)! = (1)(2)(3)...(n)(n+1)...(2n-3)(2n-2)(2n-1)(2n)$$
which is
$$(n!)(n+1)...(2n-3)(2n-2)(2n-1)(2n)$$
So I can cancel one n! term, leaving:
$$\frac{n!}{(n+1)(n+2)...(2n-2)(2n-1)(2n)}$$
Then expanding n! a little bit gives:
$$\frac{(1)(2)...(n-2)(n-1)n}{(n+1)...(2n-2)(2n-1)(2n)}$$
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 I factor out 2ⁿ from the numerator I'd have (I think!):
$$\frac{(1)(2)...(n-2)(n-1)n}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})n(2^n)}$$
Then I divide by n,
$$\frac{(1)(2)...(n-2)(n-1)}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})(2^n)}$$

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 2ⁿ 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))2ⁿ in the denominator, meaning this race to infinity is shamefully one sided (assuming my factoring of 2ⁿ isn't done incorrectly). Yes, no, maybe so? Any thoughts on a better way to do this?Thanks!

 

[Incand]

Your first solution is fine but I would express the last part a bit cleaner.
What you are doing is matching the terms one by one and note the one in the numerator is always smaller. But using this you clearly see that your expression is smaller than for example $1/(n+1)$ which has the limit zero as $n\to \infty$ so you bounded the limit above by zero while your expression is always positive.

Also check out Stirling's formula for the behaviour of $n!$ at large $n$ and you see the same result.

 

[Ray Vickson]

Homework Statement

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

Find the limit:
$$\lim_{x\rightarrow ∞}\frac{(n!)^2}{(2n)!}$$

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:

$$(2n)! = (1)(2)(3)...(n)(n+1)...(2n-3)(2n-2)(2n-1)(2n)$$
which is
$$(n!)(n+1)...(2n-3)(2n-2)(2n-1)(2n)$$
So I can cancel one n! term, leaving:
$$\frac{n!}{(n+1)(n+2)...(2n-2)(2n-1)(2n)}$$
Then expanding n! a little bit gives:
$$\frac{(1)(2)...(n-2)(n-1)n}{(n+1)...(2n-2)(2n-1)(2n)}$$
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 I factor out 2ⁿ from the numerator I'd have (I think!):
$$\frac{(1)(2)...(n-2)(n-1)n}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})n(2^n)}$$
Then I divide by n,
$$\frac{(1)(2)...(n-2)(n-1)}{(\frac{n}{2}+\frac{1}{2})...(n-1)(n-\frac{1}{2})(2^n)}$$

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 2ⁿ 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))2ⁿ in the denominator, meaning this race to infinity is shamefully one sided (assuming my factoring of 2ⁿ isn't done incorrectly).Yes, no, maybe so? Any thoughts on a better way to do this?Thanks!

Use Stirling's Formula;
http://www.sosmath.com/calculus/sequence/stirling/stirling.html
or
https://en.wikipedia.org/wiki/Stirling's_approximation

 

[micromass]

You could use the ratio test for series (yes, I know you are dealing with a sequence here!).