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

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The discussion focuses on determining the values of p for which the series ∑(1/(k log^p(k))) converges absolutely. The ratio test was initially applied, leading to the conclusion that the limit must be less than 1. An integral test was then suggested, using the function f(x) = 1/(x log^p(x)), which showed that for p in the interval (1, ∞), the series converges. Participants clarified that the base of the logarithm does not affect the convergence and emphasized the importance of starting the series at k=2 to avoid undefined terms. The final consensus is that the series converges absolutely for p greater than 1.
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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