Factorial Limits: Find the Limit of a Sequence as n Approaches Infinity

[dlf387]

Homework Statement

Find the lim as n-->inf of the sequence
{an}=
1x3x5x...x(2n-1)
_______________
n!

Homework Equations

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 believe that is converges at infinity--is this at all correct

 

[HallsofIvy]

Homework Statement

Find the lim as n-->inf of the sequence
{an}=
1x3x5x...x(2n-1)
_______________
n!

Homework Equations

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 believe that is converges at infinity--is this at all correct?

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.

 

[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

 

[Dick]

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.

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?

 

[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

 

[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

 

[Dick]

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

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

 

[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

 

[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.