# using integral test to determine convergence or divergence

• Jul 9th 2007, 03:31 PM
Possible actuary
using integral test to determine convergence or divergence
Teacher says this converges but I keep getting divergence. What am I missing or messing up on?
$\sum_{n=2}^\infty\frac{ln~n}{n^3}$

This is what I get by using integration by parts:
$\int_2^\infty\frac{ln~x}{x^3}~dx=\frac{-ln~x}{2x^2}+\frac{1}{2}\int_2^\infty\frac{dx}{x^3} =\frac{-ln~x}{2x^2}-\frac{1}{4x^2}\big|_2^\infty$

I come up with negative infinity plus a constant = infinity
• Jul 9th 2007, 03:38 PM
Jhevon
Quote:

Originally Posted by Possible actuary
Teacher says this converges but I keep getting divergence. What am I missing or messing up on?
$\sum_{n=2}^\infty\frac{ln~n}{n^3}$

This is what I get by using integration by parts:
$\int_2^\infty\frac{ln~x}{x^3}~dx=\frac{-ln~x}{2x^2}+\frac{1}{2}\int_2^\infty\frac{dx}{x^3} =\frac{-ln~x}{2x^2}-\frac{1}{4x^2}\big|_2^\infty$

I come up with negative infinity plus a constant = infinity

$\lim_{x \to \infty} \left( \frac{- \ln x}{2x^2}-\frac{1}{4x^2} \right) = 0$
• Jul 9th 2007, 03:39 PM
ThePerfectHacker
Do not use the integral test,

Note that,
$0\leq \frac{\ln n}{n^3} \leq \frac{n}{n^3} = \frac{1}{n^2}$

And $\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6} < \infty$

So it converges.
• Jul 9th 2007, 03:40 PM
Possible actuary
Quote:

Originally Posted by Jhevon
$\lim_{x \to \infty} \left( \frac{-ln~x}{2x^2}-\frac{1}{4x^2} \right) = 0$

Thanks Just now saw that the denominator in the first fraction is growing larger than ln x and therefore goes to 0 along with the second fraction.
• Jul 9th 2007, 03:41 PM
ThePerfectHacker
Quote:

Originally Posted by Possible actuary
Thanks Just now saw that the denominator in the first fraction is growing larger than ln x and therefore goes to 0 along with the second fraction.

Yes, it is useful to remember that,
$\lim_{x\to \mbox{Me}} \frac{\ln x}{x} = 0$
• Jul 9th 2007, 05:55 PM
rualin
Quote:

Originally Posted by ThePerfectHacker
Yes, it is useful to remember that,
$\lim_{x\to \mbox{Me}} \frac{\ln x}{x} = 0$

Interesting... What does the "Me" mean? Is it another joke? $TPH=\infty$?
• Jul 9th 2007, 05:56 PM
Jhevon
Quote:

Originally Posted by rualin
Interesting... What does the "Me" mean? Is it another joke? $TPH=\infty$?

yes. he does that from time to time. there are quite a few whimsical fellows on this forum... heck, i'm answering one right now, Mr. I-Own-A-Smurf :D
• Jul 10th 2007, 04:07 AM
topsquark
Quote:

Originally Posted by Jhevon
yes. he does that from time to time. there are quite a few whimsical fellows on this forum... heck, i'm answering one right now, Mr. I-Own-A-Smurf :D

I had a Smurf once. Since its skin was always blue I thought it was cold so I kept wrapping it in blankets. It died of heat prostration. :(

-Dan
• Jul 10th 2007, 04:46 AM
servantes135
lol smurfs. Hey Smurf! lets go smurf those smurfity smurf smurfs and then smurf the smurfs right out!

aint Smurfing fun!
• Jul 10th 2007, 06:39 AM
DivideBy0
Omg it's a smurf! What should I do? Stay or run?