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

1. anonymous

Are you asking if the SUM from k=1 to infinity of lnk/k^3 converges, or if the summand itself does?

2. anonymous

Use the integral test and set the integrand to ln(x)/x^3. Integrate it out to get $-\ln{x}/2{x^2}|^c_1-1/4[1/c^2-1]$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.

3. anonymous

You have to use the appropriate test to do this. There's a lot of choices, i.e. p-test, 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: $\int\limits\limits_{1}^{t}x^{-3}\ln(x)=(\ln(t)/t^{3}-1/4t^{2})-(\ln(1)/1-1/4)$ 3) Now take the lim t-> infinity: $\lim_{t \rightarrow \infty}(\ln(t)/t^{3}-1/4t^{2})-(\ln(1)/1-1/4)$$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}$