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Homework Help: 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)

    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]tpdt (from 0 to ∞)

    =[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

    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
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