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For what values of p does

  1. Sep 8, 2009 #1
    for what values of p does....

    1. The problem statement, all variables and given/known data
    Find the values of p for which the series is convergent:

    2. Relevant equations
    There are two separate problems I'm having problems with here.
    The first is:
    n=1 to n=∞ ∑ 1/(n(lnn)^p)

    The second is:
    n=1 to n=∞ ∑ ln(n)/(n^p)




    3. The attempt at a solution
    Okay, I'm almost certain both of them converge for all values p>1. I'm having trouble proving it though.

    Question 1:
    On this one, I first took the sister function
    f(x)=1/x(lnx)^p
    and integrated it using u-substitution:
    ∫ (1/x(lnx)^p )dx u=lnx du=(1/x)dx
    ∫u^(-p)du = (u^(-p+1)) / (-p+1)
    which I rewrote as:
    (u^(1-p)) / (1-p)
    then, I took the limit
    lim n→∞ (u^(1-p)) / (1-p), looking for where it would converge.

    I didn't really do any editing here, I just assumed p had to be greater than 1 so that the bottom was not negative, and not equal to 0, which would be undefined. Likewise, the top would have a negative exponent which would make the ln(x) converge. Therefore, I decided that it converges for all values p>1. I'm really not sure if I worked this one correctly or not.

    Question 2:
    I worked this one very similarly, but because it seems to be more complex, stumbled.
    Once again, I took the sister function:
    f(x)= lnx/x^p integrated by parts:
    ∫(lnx/x^p)dx u=lnx du=(1/x)dx dv=(1/x^p)dx v=x^(1-p)/(1-p)

    With this, I'm not completely sure I got the correct value for v, when integrating dv.

    However, I proceeded to set up the equation:

    uv - ∫vdu = ((lnx)(x^1-p))/(1-p) - ∫[(x^(1-p))/(x(1-p))]dx

    By this point, I didn't even bother integrating the ∫vdu because I felt so far off track.
    That's pretty much as far as I worked on that one, assuming any of it is correct.

    I'm new to this forum, so I hope I posted all of the necessary information...
    It looks a bit messy, I tried to tidy it up as well as I could with symbols, but I'm not very familiar with how to use the fractions.

    Thanks in advance for the help!!!
     
  2. jcsd
  3. Sep 8, 2009 #2

    lanedance

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    Homework Helper

    Re: for what values of p does....

    i think you mean lim u goes to infinity, which is ok as lnx goes to infinity a s x does

    to write in latex, trying clicking on text below to see the script
    [tex] \frac{u^{1-p}}{1-p} [/tex]

    you reasoning looks ok to me, but take a look at the first term in your series... may complicate things a little
    I think you're on the right track here, try simplifying the integral & keep going

    It should also be easy to do a comparision test for p<=1, with the harmonic series 1/n
     
  4. Sep 8, 2009 #3
    Re: for what values of p does....

    Okay, let's see.

    I stopped here:
    uv - ∫vdu = ((lnx)(x^1-p))/(1-p) - ∫[(x^(1-p))/(x(1-p))]dx
    and then I picked back up.

    [tex] \frac{lnx(x)^{1-p}}{1-p} [/tex] -[tex]
    \frac{1}{1-p}[/tex] ∫[tex]x^(-p)[/tex]

    = lim n→∞ [tex] \frac{ln(x)^{1-p}}{1-p} - \frac{x^{1-p}}{(1-p)^{2}} [/tex]

    At this point, I said that the limit converges for all values p > 1, because as p grows past 1, the x will be going to negative exponential powers, which converge to 0. Did I follow through well enough?

    P.S. I hope I used latex okay.
     
    Last edited: Sep 9, 2009
  5. Sep 9, 2009 #4

    lanedance

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    Homework Helper

    Re: for what values of p does....

    Another point, for any cases where p = 1, the integration steps will be different so you should consider this separately

    Back to Q2, you get to:
    [tex] \displaystyle\lim_{x\to\infty} \frac{ln(x).x^{1-p}}{1-p} - \frac{x^{1-p}}{(1-p)^{2}} [/tex]

    I think you dropped the x somewhere previously, but you will need to consider the limit of ln(x)/x^(1-p), for p > 1
     
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