pylauzier said:Homework Statement
I'm trying to self-study Mary L. Boas' book Mathematical Methods in the Physical Sciences. One of the exercices asks the reader to find the limit of n -> ∞ (n!)2 / (2n)!
Homework Equations
None
The Attempt at a Solution
Instinctively I know that (2n)! grows faster than (n!)^2, so I know the answer is zero. However I have absolutely no idea how to prove it. Can the (2n)! term be rewritten to somehow cancel out the (n!)^2 term on the numerator?
Dick said:If you want to be fancy you could use Stirlings approximation. If not, just write it out for some n. Like (1/1)*(1/2)*(2/3)*(2/4)*(3/5)*(3/6) for n=3. Does that suggest a comparison test?
Dick said:If you want to be fancy you could use Stirlings approximation. If not, just write it out for some n. Like (1/1)*(1/2)*(2/3)*(2/4)*(3/5)*(3/6) for n=3. Does that suggest a comparison test?
pylauzier said:I know how to compare sums, but I've never learned to compare sequences. What am I looking for? For example if I simplify the product given by n=3, I can see that the denominator in my sequence is bigger than the denominator of 1/n. Since lim n-> infinity of 1/n tends to zero, does that prove that lim -> infinity (n!)^2 / (2n)! will also tend towards zero?
Edit: I looked up stirling's approximation. Would I simply have to replace the n terms in the formula with 2n if I want to use it to approximate (2n)! ? Sorry for another dumb question.
Dick said:All of the terms in your series are <=1. About half of them are equal to 1/2. Keep thinking about it. Don't use Stirling's if you had to look it up. It's overkill. A comparison would be much nicer.
pylauzier said:I'm not sure I understand what I'm supposed to compare. I thought the comparison test was for series (sums of terms), not simple limits. If I'm not asking for too much, would you care pointing me towards the right direction? I even searched in my old calc II book and there is no mention of a comparison test for sequences (factorials). :(
pylauzier said:Ok, I think I understand now. The nth term will be smaller than (1/2^n), since all the terms are smaller than one and because there is n 1/2 terms.
So lim n -> ∞ (n!)2 / (2n)! =< lim n -> ∞ (1/2^n)
Since lim n -> ∞ (1/2^n) = 0, it follows that
lim n -> ∞ (n!)2 / (2n)! =< 0
Dick said:Sure, that's it!
A quotient of factorials is a mathematical expression that involves dividing one factorial (n!) by another factorial (m!). It is represented as n!/m!.
Simplifying a quotient of factorials can help make the expression easier to work with and understand. It can also help to identify patterns and relationships between different factorials.
To simplify a quotient of factorials, you can use the properties of factorials, such as n! = n*(n-1)!, to expand and cancel out common factors. You can also use the division rule of factorials, which states that n!/m! = (n*(n-1)*...*(m+1)), to simplify the expression further.
No, not all quotients of factorials can be simplified. Some expressions may not have any common factors to cancel out, or may not follow the division rule of factorials. In these cases, the quotient of factorials cannot be simplified any further.
Some common mistakes when simplifying a quotient of factorials include not using the correct properties of factorials, forgetting to distribute the factorial to all terms, and not simplifying the expression fully. It is important to double check your work and make sure all steps are correct when simplifying a quotient of factorials.