# Converges absolutely, conditionally, or diverges

1. Jan 21, 2008

### rcmango

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

does this series converge absolutely, conditionally, or diverge?

what test to use for this problem: http://img525.imageshack.us/img525/2296/35297847wb1.png [Broken]
2. Relevant equations

3. The attempt at a solution

not sure where to start. also n!^2 is that a p-series?

thankyou.

Last edited by a moderator: May 3, 2017
2. Jan 21, 2008

### arildno

Argue, that for large enough n, (n+2)!/(n!)^2<1

3. Jan 21, 2008

### rcmango

is this a series that I could use the limit comparison test, so compare that to 1? then compare this to the original problem?

also, looks like a p-series to me.
Thanks so far.

4. Jan 21, 2008

### dynamicsolo

That's pretty definitely not a p-series. Such a series would simply use n raised to a power, rather than n!

Have you had the Ratio Test yet? That's really the most effective technique on a general term of this sort. Also, if you succeed in showing that this series converges absolutely (since it contains no alternating term), you'll be done. (Ah, but perhaps I've said too much...)

(Hmm, if you have only gotten to the comparison tests, you'd need to use a ratio bigger than the general term that you know will give a convergent series. You could make the denominator smaller by looking at the series for just (n+2)!/[ (3^n) ยท n! ] , simplify that , and show that this series converges, so yours does, too.)

Last edited: Jan 21, 2008
5. Jan 21, 2008

### arildno

Note that:
$$\frac{(n+2)!}{(n!)^{2}}=\frac{n!}{n!}\frac{(n+1)(n+2)}{n!}=\frac{(1+\frac{1}{n})(1+\frac{2}{n})}{(n-1)!}$$

What does this tell you?

6. Jan 22, 2008

### rcmango

Thanks for the help. I may need help simplifying the factorial in the denominator though.

i believe that now looks like 1/ (n -1)! ?

where 1/n = 0 in and 2/n = 0. so 1/ the factorial is less than one, correct?

what happened to the 3n?

thankyou.

7. Jan 22, 2008

### arildno

What happened to it?

It is still there!

What you have shown is that given n>3, the terms in the series is less than 1/3^n.

What does that tell you about the series?