# Sequence homework problem help please

1. Jul 30, 2007

### Winzer

1. The problem statement, all variables and given/known data
Determine wheatehr the sequence diiverges or converges:

2. Relevant equations
$$a_{n}=\frac{(n+2)!}{n!}$$

3. The attempt at a solution
I was gonna treat it using limits but the factorial is not defined for a function.
How do I deal with this?
Edit: sorry it is suppose to be: $$a_{n}=\frac{(n+2)!}{n!}$$ not$$a_{n}=\frac{(n+1)!}{n!}$$

Last edited: Jul 30, 2007
2. Jul 30, 2007

### CaptainZappo

Try rewriting (n+1)! in a different form. What do you notice about the relationship between (n+1)! and n!?

3. Jul 30, 2007

### Winzer

Ok I am pretty sure the it will diverge because (n+1)! goes faster to infinity than n!

rewrite like: $$a_{n}= 1+\frac{1}{n!}$$ ?

4. Jul 30, 2007

### CaptainZappo

Your rewritten version of the equation is not equal to the original equation. In fact, I believe the rewritten equation converges to 1 (do you see why?), thereby contradicting your statement above.

Please rewrite the numerator again, keeping in mind the goal of trying to simplify the equation.

5. Jul 30, 2007

### Winzer

can i say n! = n(n-1)! and sub that in?

6. Jul 30, 2007

### CaptainZappo

Yes! You are on the right track. If you try rewriting the numerator instead of the denominator, I think you will have more luck.

7. Jul 30, 2007

### Winzer

but how would i rewrite the numerator?

8. Jul 30, 2007

### matt grime

So, you know that n!= n(n-1)!... why is it not clear to you that that is precisely the same as (n+1)!=(n+1)n!? Or (n+1)!=(n+2)*(n+1)! ?????

9. Jul 30, 2007

### Winzer

how is n!= n(n-1)! the same as (n+1)!=(n+1)n!??

10. Jul 30, 2007

### Winzer

what I meant was how did u get that?

11. Jul 31, 2007

### Dick

The formula is true for all n. If n!=n*(n-1)! then changing n->n+1 gives (n+1)!=(n+1)*n!. In this sense, it's the 'same formula'. You may wish to try this for n->n+2. Can you show (n+2)!=(n+2)*(n+1)*n! at least for large enough n?

12. Jul 31, 2007

### Winzer

Oh ok, thanks dick for clarifying, I just didn't think of it like that.
Indeed I worked it out and n->n+2 is (n+2)!=(n+2)*(n+1)*n!.