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

• Battlemage!
In summary: But the ratio test is essentially the same as what you did with the n's in the denominator. (You might need to use the absolute value of the ratio to be sure it works. Then you need to explain why the resulting terms go to 0.)
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!

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.

Battlemage!
Battlemage! said:

## 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

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

Battlemage!

## 1. What is the significance of the limit as n approaches infinity?

The limit as n approaches infinity is an important concept in calculus and mathematical analysis. It represents the behavior of a function at its "endpoints" and helps us understand the overall behavior and properties of the function.

## 2. How does the expression n^2/(2n) represent the limit as n approaches infinity?

The expression n^2/(2n) can be simplified to 1/2n, which represents the limit as n approaches infinity. This means that as n gets larger and larger, the value of the expression approaches 0. This behavior is characteristic of a function with a horizontal asymptote at y=0.

## 3. Why is this reasoning considered acceptable for the limit as n approaches infinity?

This reasoning is considered acceptable because it follows the fundamental theorem of calculus, which states that the limit of a ratio of two functions is equal to the ratio of their limits (as long as the denominator does not approach 0). In this case, as n approaches infinity, both the numerator and denominator approach infinity, resulting in a limit of 1/2.

## 4. Are there any other ways to represent the limit as n approaches infinity?

Yes, there are multiple ways to represent the limit as n approaches infinity. Some common notations include lim n->∞ f(n), f(n)->∞ as n->∞, and f(n)→∞ as n→∞. All of these notations represent the concept of a function approaching a value or behavior as the input (n) becomes infinitely large.

## 5. Is the concept of a limit as n approaches infinity limited to only rational expressions?

No, the concept of a limit as n approaches infinity can be applied to any type of function, including irrational, trigonometric, and logarithmic functions. However, in order for this concept to be useful, the function must have a well-defined limit as n approaches infinity. Some functions may not have a limit at infinity, and therefore the concept of a limit would not apply.

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