complicated p-series

• Mar 22nd 2009, 01:00 PM
buttonbear
complicated p-series
find the values of p for which the series is convergent:

$\displaystyle \sum_{n=3}^{\infty}\frac{1}{n ln n [ln(ln n)]^p}$

• Mar 22nd 2009, 01:10 PM
skeeter
integral test ...

$\displaystyle \int_3^{\infty} \frac{1}{x\ln{x}[\ln(\ln{x})]^p} \, dx$

$\displaystyle u = \ln{x}$

$\displaystyle du = \frac{1}{x} \, dx$

$\displaystyle \int_{\ln{3}}^{\infty} \frac{1}{u(\ln{u})^p} \, du$

consider three cases ... p < 1, p = 1, and p > 1
• Mar 22nd 2009, 01:23 PM
buttonbear
i'm kind of confused because i thought that the integral test was supposed to test for convergence or divergence..but don't you already know it will be convergent somewhere, because it's a p-series? i thought that you couldn't use the evaluation of the integral to tell you where the series actually converges to.. so, should i just evaluate the limit of the improper integral (once i've integrated) and then.. how do the p values i'm supposed to consider come into play? (Thinking)
• Mar 22nd 2009, 01:28 PM
skeeter
Quote:

Originally Posted by buttonbear
i'm kind of confused because i thought that the integral test was supposed to test for convergence or divergence..but don't you already know it will be convergent somewhere, because it's a p-series? i thought that you couldn't use the evaluation of the integral to tell you where the series actually converges to.. so, should i just evaluate the limit of the improper integral (once i've integrated) and then.. how do the p values i'm supposed to consider come into play? (Thinking)

the "p" in this problem is not the same as the "p" in the series $\displaystyle \sum{\frac{1}{n^p}}$ ... it's just a variable for the exponent.
• Mar 22nd 2009, 01:31 PM
buttonbear
so, how can you integrate 1/u(ln u)^p with the two variables? i'm just kind of lost..
• Mar 22nd 2009, 01:49 PM
skeeter
Quote:

Originally Posted by buttonbear
so, how can you integrate 1/u(ln u)^p with the two variables? i'm just kind of lost..

$\displaystyle \int \frac{1}{u (\ln{u})^p} \, du$

remember substitution?

$\displaystyle t = \ln{u}$

$\displaystyle dt = \frac{1}{u} \, du$

$\displaystyle \int \frac{1}{t^p} \, dt$
• Mar 22nd 2009, 01:51 PM
buttonbear
honestly, i can't claim i understand any of these 'tools' enough to be able to wield them for this kind of complicated stuff.. thanks for your help(Shake)
• Mar 22nd 2009, 03:40 PM
buttonbear
if p=1 then i know it just comes out to be ln t...but if it's >1 or < 1 then it's the power rule... so how do you test if it's convergent? i know it's not right to just come on here and ask for answers, but i'm kind of struggling, do you think you could walk me through it?
• Mar 22nd 2009, 04:07 PM
skeeter
$\displaystyle \int \frac{1}{t^p} \, dt$ converges for $\displaystyle p > 1$, diverges for $\displaystyle p \leq 1$
• Mar 22nd 2009, 05:04 PM
buttonbear
thank you so much for your help

this seems like an awful lot of work if this function/sequence behaves exactly the same way as the p-series(Shake)