# Another Question for a P-Series Problem

• Sep 10th 2012, 09:29 AM
Beevo
Another Question for a P-Series Problem
Determine all values of p for which the series is convergent, and express answer in interval notation.

$\displaystyle \sum_{n=2}^{\infty} (-1)^{n-1} \frac{ln(n)}{2n}^{p}$

How would I find the values of p that makes this series a convergent one. It appears to be an alternating series, so would I just focus on $\displaystyle \frac{ln(n)}{2n}^{p}$ ?
I am just not quite sure how to go about this problem. Could I used the ratio test or anything else?

Any feedback and help appreciated, thanks.
• Sep 10th 2012, 10:45 AM
MaxJasper
Re: Another Question for a P-Series Problem
Quote:

Originally Posted by Beevo
Determine all values of p for which the series is convergent, and express answer in interval notation.
$\displaystyle \sum_{n=2}^{\infty} (-1)^{n-1} \frac{ln(n)}{2n}^{p}$

For p=1000 the series converges to $\displaystyle 4.312152513533860\times 10^{554}$

It appears that the series converges for all $\displaystyle p<\infty$

http://mathhelpforum.com/attachment....1&d=1347305026
• Sep 10th 2012, 12:20 PM
skeeter
Re: Another Question for a P-Series Problem
$\displaystyle \frac{\ln{n}^p}{2n} = \frac{p}{2} \cdot \frac{\ln{n}}{n}$

since $\displaystyle \lim_{n \to \infty} \frac{\ln{n}}{n} = 0$ , and the series is alternating ...

$\displaystyle \sum_{n=2}^{\infty} (-1)^{n-1}\frac{\ln{n}^p}{2n} = \frac{p}{2} \sum_{n=2}^{\infty} (-1)^{n-1} \frac{\ln{n}}{n}$

... is conditionally convergent for all fixed values of p.
• Sep 10th 2012, 05:17 PM
Beevo
Re: Another Question for a P-Series Problem
Quote:

Originally Posted by skeeter
$\displaystyle \frac{\ln{n}^p}{2n} = \frac{p}{2} \cdot \frac{\ln{n}}{n}$

since $\displaystyle \lim_{n \to \infty} \frac{\ln{n}}{n} = 0$ , and the series is alternating ...

$\displaystyle \sum_{n=2}^{\infty} (-1)^{n-1}\frac{\ln{n}^p}{2n} = \frac{p}{2} \sum_{n=2}^{\infty} (-1)^{n-1} \frac{\ln{n}}{n}$

... is conditionally convergent for all fixed values of p.

I came to this same conclusion. Since it is an alternating series, I took the limit of bn and found that it was converging.

However, there is also the rule that the series has to be decreasing for it to be convergent in a alternating series. So, would the value of p affect the rule of decrease? That is what I am stuck on. If not, then clearly the interval notation for the value of p that makes this series converge is [o to infinity).
• Sep 10th 2012, 05:23 PM
skeeter
Re: Another Question for a P-Series Problem
note that p/2 is a constant that can be factored out of the series terms ...

$\displaystyle k \sum b_n$ where k is a constant is convergent if $\displaystyle \sum b_n$ is convergent.
• Sep 10th 2012, 05:28 PM
Beevo
Re: Another Question for a P-Series Problem
Quote:

Originally Posted by skeeter
note that p/2 is a constant that can be factored out of the series terms ...

$\displaystyle k \sum b_n$ where k is a constant is convergent if $\displaystyle \sum b_n$ is convergent.

Got it!!! Thanks man.

I did not realize that the p could be factored out to form a constant. Makes sense now.
• Sep 10th 2012, 05:29 PM
MaxJasper
Re: Another Question for a P-Series Problem
I suspect the correct term in the series might be: $\displaystyle [\ln (n)]^p$
• Sep 10th 2012, 05:32 PM
Beevo
Re: Another Question for a P-Series Problem
Quote:

Originally Posted by MaxJasper
I suspect the correct term in the series might be: $\displaystyle [\ln (n)]^p$

Yeah, the entire natural log function is raised to the p. Have I been solving it incorrectly?
• Sep 10th 2012, 05:35 PM
MaxJasper
Re: Another Question for a P-Series Problem
Others assumed ln[n^p]
• Sep 10th 2012, 05:39 PM
Beevo
Re: Another Question for a P-Series Problem
Oh really? That's a problem. Then I have been solving it incorrectly. Well it is still an alternating series, I just need to figure out what values of p make this series convergent. Any ideas?