1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Am i right here?

  1. Apr 16, 2009 #1
    i have reached a possible answer which is not quite like the one my book has, can someone please tell me if im wrong.

    the question asks if the following converges or diverges.


    [tex]\sum[/tex](-1)n-1(ln(n))p/n (p>0)
    n=1

    i learned that if i have a series which changes sign i take the abs value of [tex]\sum[/tex]An and test that series, if [tex]\sum[/tex]|An| converges then An definitely converges, if [tex]\sum[/tex]|An| diverges, i need to check if lim(n->∞)An=0 and if An>A(n+1) (leibnitz)
    i integrated the abs value of function in order to find its behaviour, if i get ∞ then the abs series diverges

    [tex]\int[/tex]((ln(x))p/x)dx
    t=ln(x)
    dt=dx/x
    x=et

    [tex]\int[/tex]tpdt (from 0 to ∞)

    =[lim[b->∞]][tex]\frac{1}{p+1}[/tex](tp+1)|[tex]^{b}_{0}[/tex]
    =[lim[b->∞]][tex]\frac{1}{p+1}[/tex](bp+1-0p+1)= ∞
    therefore the abs series diverges so i need to check limAn and An+1

    lim An = lnp(n)/n
    n->∞

    surely this is dependant on p, i need to be equal to 0, therefore i need n to be bigger than lnp(n),(since we are talking about a limit of n-> ∞)

    lnp(n) < n
    p < logln(n)(ln(n))

    is this correct, or will lnp(n) < n always be true since n->∞???

    now how do i prove than An > An+1

    [tex]\frac{lnp(n+1)}{n+1}[/tex] < [tex]\frac{lnp(n)}{n}[/tex]
    ?????

    alternatively, do you see any better ways to solve this??
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?