Factorial Ratio for Interval of Convergence Calculation

[queensu]

Homework Statement

Hi! I need to find ratio (2n+1)!/(2n+3)! for interval of convergence calculation.

Homework Equations

5! = 1*2*3*4*5

The Attempt at a Solution

i have no idea where to start since i have never dealt with factorials before.. if you just show me some kind off factorial property i can use ill be able to figure it out. wikipedia didnt help.

 

[Bohrok]

5! = 5*4*3*2*1 and 4! = 4*3*2*1, so
5! = 5*4!
5! = 5*4*3!

n! = n*(n -1)!
n! = n*(n-1)*(n-2)!
and more generally, n! = n*(n-1)*(n-2)...3*2*1

See if you can break down (2n + 1)! and (2n + 3)! so you can cancel something.

 

[queensu]

oh.. awesome thanks a lot.
after very long thing of expanding that i got 1/ (2n+3)

 

[JCVD]

Your answer is not quite right, and obtaining the correct one should not take much expanding. You only need to apply the identity n! = n*(n-1)*(n-2)! to the denominator.

 

[queensu]

(2n+3)=(2n+3)*(2(n-1)+1)! = 2n+3)*(2n-2+3)! =(2n+3)(2n+1)! ||| (2n+1)! cancels and I am left with 1/(2n+3)

 

[Bohrok]

You should have (2n+2) somewhere; looks like you factored out the 2 and then it got lost...

 

[dimension10]

Homework Statement

Hi! I need to find ratio (2n+1)!/(2n+3)! for interval of convergence calculation.

Homework Equations

5! = 1*2*3*4*5

The Attempt at a Solution

i have no idea where to start since i have never dealt with factorials before.. if you just show me some kind off factorial property i can use ill be able to figure it out. wikipedia didnt help.

I think it is 1/(4n^2+10n+6)

 

[Saitama]

Like Bohrok said,

5! = 5*4*3*2*1 and 4! = 4*3*2*1, so
5! = 5*4!
5! = 5*4*3!

Use this logic to simplify your question.
It would go like this:-

$$\frac{(2n+1)!}{(2n+3)(2n+2)(2n+1)!}$$

Now cancel out (2n+1)! and you get:-

$$\frac{1}{(2n+3)(2n+2)}$$

Done !

 

[dimension10]

Like Bohrok said,

5! = 5*4*3*2*1 and 4! = 4*3*2*1, so
5! = 5*4!
5! = 5*4*3!

Use this logic to simplify your question.
It would go like this:-

$$\frac{(2n+1)!}{(2n+3)(2n+2)(2n+1)!}$$

Now cancel out (2n+1)! and you get:-

$$\frac{1}{(2n+3)(2n+2)}$$

Done !

Umm... that can be simplified to 1/(4n^2+10n+6). From this, we can see that when and only when n=-1 or -1.5, (2n+1)!/(2n+3)!=infinity

 

[Saitama]

What did queensu mean by "for interval of convergence calculation"?