Find p in R for Absolute Conv. Series: 1/(klog^pk)

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Homework Help Overview

The discussion revolves around determining the values of p in the real numbers for which the series \(\sum_{k=2}^{\infty} \frac{1}{k \log^p k}\) converges absolutely. The subject area includes series convergence and the application of convergence tests.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the ratio test and integral test to analyze the convergence of the series. There are attempts to manipulate the series and integral expressions to explore convergence conditions, particularly focusing on the behavior of the logarithmic terms as k approaches infinity.

Discussion Status

Some participants have provided guidance on using the integral test and have shared their reasoning regarding the limits involved. There is an ongoing exploration of the implications of different values of p on convergence, with no explicit consensus reached yet.

Contextual Notes

Participants note the importance of the lower limit of the series starting at k=2 to avoid undefined terms. There is also a discussion about the relevance of absolute values in the context of the function being positive and decreasing.

wany
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Homework Statement


find all values of p in R for which the given series converges absolutely.
\displaystyle\sum\limits_{k=2}^{\infty} \frac{1}{klog^pk}

Homework Equations


Ratio Test Or Root Test

The Attempt at a Solution


So I tried using the ratio test and get:
\mathop {\lim }\limits_{k \to \infty } \frac{|a_{k+1}|}{|a_k|}=\mathop {\lim }\limits_{k \to \infty } |\frac{klog^pk}{(k+1)log^p(k+1)}|=\mathop {\lim }\limits_{k \to \infty } |\frac{log^pk}{log^p(k+1)}| since lim as k goes to infinity of k/(k+1) is 1.
We know that we want this limit to be less than 1.

I am stuck from this point. Any help would be appreciated (there should be absolute signs in there I am not sure why they didnt show up).
 
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Use an integral test.
 
ok, so i let f(x)=\frac{1}{xlog^px} and then when I take the integral of this function, first I let log^px=\frac{lnx^p}{ln10^p} and then I let u=lnx
so I get that the integral is ln10^p*\frac{-u^{1-p}}{p-1} taken from 1 to t and taking the limit as t goes to infinity, so ln10^p*\frac{ln(1)^{1-p}-lnt^{1-p}}{p-1}
So we see that when p is in (1, infinity), that makes sure that lnt^{1-p} goes to 0 instead of infinity and thus our series converges. Is this correct? And thank you for your help so far.
 
wany said:
ok, so i let f(x)=\frac{1}{xlog^px} and then when I take the integral of this function, first I let log^px=\frac{lnx^p}{ln10^p} and then I let u=lnx
so I get that the integral is ln10^p*\frac{-u^{1-p}}{p-1} taken from 1 to t and taking the limit as t goes to infinity, so ln10^p*\frac{ln(1)^{1-p}-lnt^{1-p}}{p-1}
So we see that when p is in (1, infinity), that makes sure that lnt^{1-p} goes to 0 instead of infinity and thus our series converges. Is this correct? And thank you for your help so far.

I think that's actually pretty much it. I wouldn't worry about the logs base 10 vs ln. I think you can take log=ln if it doesn't specifically say log_10. And they did set the lower limit of the sum to n=2 for a good reason. Otherwise the first term in the sequence is undefined. But other than that it looks ok to me.
 
Oh you it is 2 and not 1. Thank you for your help, I appreciate it.
 
One more thing, since I want it to converge absolutely, do I need to have the integral of the absolute function?
 
Never mind that was a stupid question, the function is positive and decreasing, so absolute does not matter in this case.
 

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