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lorgarithmic p-series question

skatenerd

Active member
Oct 3, 2012
114
So I've been trying to make some progress on this question for my Calc 2 class, here it is:
"The Logarithmic p-series is defined by (sigma summation n=2 to infinity) of 1/(n(ln(n)))p) for p>0. Determine for which values of p it is convergent or divergent."

So off the bat I kind of assumed that to do this with a constant p, and seeing the series is decreasing and positive, I should use the integral test. Set up the integral of this series, and I made a substitution u for ln(n). this ended up with (u-p+1)/(p+1) and I plugged the ln(n) back in and this is where I got kind of lost. Plugging in the limits of integration seemed to be a little messy, and I didn't really know where to go from there. If someone wants to give me a hint or a little guidance that would be nice. Thanks.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
I just replied to this but deleted it because I misinterpreted your problem. I would suggest taking some time to learn Latex as it makes things much easier to read.

So you have \(\displaystyle \sum_{n=2}^{\infty}\frac{1}{n \left(\ln n \right)^p}\)

As you said with the substitution of $u=\ln n$ this integral becomes \(\displaystyle \int \frac{1}{u^p}du=u^{-p}du\) We can solve this and then back-substitute, then plug in the limits.

You were close on your solution, but the denominator should be -p, not p.

\(\displaystyle \int u^{-p}du=\frac{u^{-p+1}}{-p+1}+C\)

Plugging back in the substitution $u=\ln n$ that becomes

\(\displaystyle \frac{(\ln n)^{-p+1}}{-p+1}\)

That's pretty much where you are though, so how do we proceed from here? Consider the end behavior of $\ln n$ as $n \rightarrow \infty$. Now also consider positive and negative exponent rules. What happens when there is a negative exponent?

More specifically, if the exponent in the numerator $-p+1$ is positive then does this converge or diverge? What about if it's negative?
 
Last edited:

skatenerd

Active member
Oct 3, 2012
114
(Having a hard time finding latex...)
Okay that helped a lot. At first your questions were kind of getting me confused, but I wrote it out and I think I got the answer. when 0<p<1, the numerator will approach infinity and therefore the whole term diverges. And when p>1 the numerator will have to be flipped over and multiplied by 1/-p+1. Therefore the whole term will always approach zero and converge. Correct?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
(Having a hard time finding latex...)
Okay that helped a lot. At first your questions were kind of getting me confused, but I wrote it out and I think I got the answer. when 0<p<1, the numerator will approach infinity and therefore the whole term diverges. And when p>1 the numerator will have to be flipped over and multiplied by 1/-p+1. Therefore the whole term will always approach zero and converge. Correct?
Here is our forum for Latex.

Yep, you're spot on. For $p<1$ (it doesn't just have to be between 0 and 1 in theory) the exponent is positive and the integral diverges, thus the series diverges. When $p>1$ the exponent flips and the $\ln n$ term is now in the denominator so the integral converges. When $p=1$ the integral diverges just because we are dividing by $-p+1$ so the whole thing converges for $p>1$, which should be pointed out is different than $p \ge 1$.

I think you got all of that already but just want to double check.
 

skatenerd

Active member
Oct 3, 2012
114
Yep. That makes perfect sense now. Thanks! I appreciate the help.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Just an addendum, it has to be $0<p<1$ because one of the hypothesis is $p>0$. I believe that if you take out that it would still be valid, but it's best to remember that bit. :D
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
Just an addendum, it has to be $0<p<1$ because one of the hypothesis is $p>0$. I believe that if you take out that it would still be valid, but it's best to remember that bit. :D
Oops. Missed that. Good catch :)