complicated p-series

**buttonbear** · March 22nd 2009, 02:00 PM

find the values of p for which the series is convergent:

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

please help!

**skeeter** · March 22nd 2009, 02:10 PM

integral test ...

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

$u = \ln{x}$

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

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

consider three cases ... p < 1, p = 1, and p > 1

**buttonbear** · March 22nd 2009, 02:23 PM

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?

**skeeter** · March 22nd 2009, 02:28 PM

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?

the "p" in this problem is not the same as the "p" in the series $\sum{\frac{1}{n^p}}$ ... it's just a variable for the exponent.

**buttonbear** · March 22nd 2009, 02:31 PM

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

**skeeter** · March 22nd 2009, 02:49 PM

Originally Posted by buttonbear

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

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

remember substitution?

$t = \ln{u}$

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

$\int \frac{1}{t^p} \, dt<br />$

**buttonbear** · March 22nd 2009, 02:51 PM

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

**buttonbear** · March 22nd 2009, 04:40 PM

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?

**skeeter** · March 22nd 2009, 05:07 PM

$\int \frac{1}{t^p} \, dt$ converges for $p > 1$ , diverges for $p \leq 1$

**buttonbear** · March 22nd 2009, 06:04 PM

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

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