# Direct Comparison test of 1/n! to what?

• Dec 19th 2012, 06:48 PM
MSUMathStdnt
Direct Comparison test of 1/n! to what?
What should I compare $\sum_{n \ge 0} \frac 1{n!}$ if the question requires me to prove its convergence/divergence using the direct comparison test. I know it converges. It seems like $\frac 1{n!} < \frac 1{n^2}$ for $n \ge 4$, and $\frac 1{n^2}$ converges by p-series test. The first four terms are obviously finite. Is that enough? Is there a better series to compare it to?
• Dec 19th 2012, 09:08 PM
chiro
Re: Direct Comparison test of 1/n! to what?
Hey MSUMathStdnt.

The p-series comparison should be more than sufficient.
• Dec 19th 2012, 10:12 PM
MSUMathStdnt
Re: Direct Comparison test of 1/n! to what?
How do you prove that 1/n! < 1/n^2?
• Dec 19th 2012, 10:59 PM
chiro
Re: Direct Comparison test of 1/n! to what?
Prove n! > n^2 for all appropriate n (i.e. n > 3).
• Dec 20th 2012, 12:33 AM
Deveno
Re: Direct Comparison test of 1/n! to what?
which is the same as proving (n-1)! > n. induction is your friend, here.

or you can just use "brute force":

(n-1)! > (n-1)(n-2) = n2 - 3n + 2 > n2 - 3n = n(n - 3) ≥ n, if n ≥ 4 (so that n - 3 ≥ 1).
• Dec 20th 2012, 07:27 AM
hollywood
Re: Direct Comparison test of 1/n! to what?
Or, what amounts to the same thing as Deveno did, if $n \ge 4$,

$n! \ge n(n-1)(n-2) = n^3 - 3n^2 + 2n > n^3 - 3n^2 = n^2(n-3) \ge n^2$.

- Hollywood
• Dec 20th 2012, 03:04 PM
MSUMathStdnt
Re: Direct Comparison test of 1/n! to what?
How come some threads have a little "thanks" link in the posts of people who reply to you and some (like this one) don't?
P.S. Thanks for replies, guys (and gals, if any).
• Dec 20th 2012, 08:15 PM
hollywood
Re: Direct Comparison test of 1/n! to what?
Some guesses:

1. You need to be logged in - in fact, I think you need to be logged in when you download the page.

2. You can not give thanks to yourself.

3. You have to mouse over the post.

It's working for me on this thread now - I've never noticed that it's not working.

- Hollywood