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anonymous
 5 years ago
Does (lnk)/(k^3), where as k starts at one and continues on to infinity, converge?
anonymous
 5 years ago
Does (lnk)/(k^3), where as k starts at one and continues on to infinity, converge?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you asking if the SUM from k=1 to infinity of lnk/k^3 converges, or if the summand itself does?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use the integral test and set the integrand to ln(x)/x^3. Integrate it out to get \[\ln{x}/2{x^2}^c_11/4[1/c^21]\]where the limits of integration are 1 to c. Next take the limit as c goes to infinity, and use L'Hopital's Rule on ln(c)/(2c^2) as c goes to infinity. In the end you should end up with 1/4. Since the integral converges, so does the sum.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to use the appropriate test to do this. There's a lot of choices, i.e. ptest, divergence test, ratio test, root test, comparison test, etc., but the best bet is the integral test which can only be done on a continuous, positive, decreasing function in the interval [k,infinity) which this series is. So, here we go: 1) First we have to establish whether \[\ln(k)/k^{3}\]is positive and decreases eventually and to do this, we check the derivative: \[d/dx(\ln(x)/x^{3})=3x^{4}\ln(x)+x^{3}(1/x)=(3\ln(x)+1)/x^{4}\]so,\[f'(x)=(3\ln(x)+1)/x^{4}\]has a critical point at \[x=e^{1/3}\]and if you pick a couple points you'll notice that the derivative is positive (increasing) from [1,e^{1/3}] and negative (decreasing) from [e^{1/3},infinity) This confirms that we can use the integral test. 2) Construct the improper integral\[\int\limits_{1}^{\infty}x^{3}\ln(x)=x^{3}\ln(x)(x^{2}/4)\]This is done using integration by parts and selecting\[u=\ln(x), dv=x^{3}dx\] Now, apply the integration limits, replacing infinity with t: \[\int\limits\limits_{1}^{t}x^{3}\ln(x)=(\ln(t)/t^{3}1/4t^{2})(\ln(1)/11/4)\] 3) Now take the lim t> infinity: \[\lim_{t \rightarrow \infty}(\ln(t)/t^{3}1/4t^{2})(\ln(1)/11/4)\]\[1/3t^{3}1/4t^{2}0+1/4=1/4\]using L'Hospitals rule on the first term. 4)Since the improper integral of the series converges to 1/4 so does the series:\[\sum_{k=1}^{\infty}\ln(k)/k^{3}\]
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