# Limits of factorials

1. Dec 31, 2009

### dlf387

1. The problem statement, all variables and given/known data
Find the lim as n-->inf of the sequence
{an}=
1x3x5x...x(2n-1)
_______________
n!

2. Relevant equations

3. The attempt at a solution
I rewrote it as
...(2n-3)(2n-2)(2n-1)
__________________
n(n-1)(n-2)...2x1
which leads me to belive that is converges at infinity--is this at all correct

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Dec 31, 2009
2. Dec 31, 2009

### HallsofIvy

No, that's not correct. And I don't know why you would say that leads you to believe it converges. Obviously, the the denominator cancels with part of the numerator:
$$\frac{(2n-1)!}{n!}= (n+1)(n+2)\cdot\cdot\cdot(2n-1)$$

That does not converge.

3. Dec 31, 2009

### dlf387

Thank you for your reply but I have not had much experience with factorials. Could you please fill in the steps to reach how you got to that final expression? Also, could you please help me understand how to get a finite limit from a factorial?
thanks

4. Dec 31, 2009

### Dick

I think the question is actually about (2n-1)!!/n!. I.e. for n=5, (1*3*5*7*9)/(1*2*3*4*5). If you write that as (1/1)*(3/2)*(5/3)*(7/4)*(9/5) you can see that all of the factors are greater than 1. In fact, a lot of them are greater than 1.5. If n is large what does this tell you about the product?

5. Dec 31, 2009

### dlf387

Thank you very much for the reply. When you write it out like that I can see that the product will eventually reach infinity. But how did you know to rewrite the expression as (2n-1)!!/n!
and what does a double factorial mean?
thanks

6. Dec 31, 2009

### dlf387

Thank you very much for the reply. When you write it out like that I can see that the product will eventually reach infinity. But how did you know to rewrite the expression as (2n-1)!!/n!
and what does a double factorial mean?
thanks

7. Dec 31, 2009

### Dick

You can look up 'double factorial' on line. It's just a shorthand way of writing your product, 1*3*5*7*9=9!!.

8. Dec 31, 2009

### dlf387

I am sorry to belabor the point, but is there anyway to reduce (2n-1)!!/n!
to a more simpler form--like the 1st commentator did? is what they did correct?

i think that n! can be written as (1x2x(n-1)x(n-2)x...x(n)) but could that cancel with anything from the numerator?
thanks

9. Dec 31, 2009

### Dick

Not really. The 2n-1 in the numerator doesn't cancel with anything in the denominator and the even numbers in the denominator don't cancel with anything in the numerator.