What value of p does the following series converge

A Theorem say the $\displaystyle p$-series $\displaystyle \sum_{n=2}^\infty \frac {1}{n^p}$ converges if $\displaystyle p>1$ and diverges otherwise.

The question says "Find the positive values of $\displaystyle p$ for which the series $\displaystyle \sum_{n=2}^\infty \frac 1{n (\ln n)^p}$ converges.

The answer is $\displaystyle p=1$ (from back of book). There is no explanation of how they got it and I have none either. *How do you find that?*

I thought maybe I would compare $\displaystyle \frac 1{n (\ln n)^p}$ to $\displaystyle \frac {1}{n^p}$. If $\displaystyle \frac 1{(n \ln n)^p} < \frac {1}{n^p}$, then the former converges, too. For $\displaystyle p>1$, and since $\displaystyle n>1$, then $\displaystyle (\ln n)^p>1 $and thus $\displaystyle \frac 1{(n \ln n)^p} < \frac {1}{n^p}$. How am I doing so far?

But what about when $\displaystyle p<1$? Maybe for some values close to, but less than 1, $\displaystyle \sum_{n=2}^\infty \frac 1{n (\ln n)^p}$ still converges. How do I find out?

Re: What value of p does the following series converge

Hey MSUMathStdnt.

Have you tried using some other tests like the integral test?

Re: What value of p does the following series converge

Quote:

Originally Posted by

**MSUMathStdnt** A Theorem say the $\displaystyle p$-series $\displaystyle \sum_{n=1}^\infty \frac {1}{n^p}$ converges if $\displaystyle p>1$ and diverges otherwise.

The question says "Find the positive values of $\displaystyle p$ for which the series $\displaystyle \sum_{n=2}^\infty \frac 1{n (\ln n)^p}$ converges.

The answer is $\displaystyle p=1$ (from back of book). There is no explanation of how they got it and I have none either. *How do you find that?*

There is a true difficulty with your post.

$\displaystyle \sum_{n=2}^\infty \frac 1{n (\ln n)}$ diverges. $\displaystyle p=1$

Look at the integral test.

So what is your question now?

Re: What value of p does the following series converge

First a correction. The lower bounds was supposed to be $\displaystyle 2$, not $\displaystyle 1$. If you try to take the $\displaystyle \ln(1)$ in the denominator... well, you know.

OK, so here's the integral test (so far)

$\displaystyle \int_2^\infty \frac 1{n (\ln n)^p} dn = \frac 1{(1-p)(\ln n)^p} \bigg|_1^\infty$

The first term

$\displaystyle \lim_{n \to \infty} \frac 1{(1-p) (\ln n)^p} = 0$ for all positive values of $\displaystyle p \ne 1$.

The second term

$\displaystyle \lim_{n \to 2^+} \frac 1{(1-p) (\ln n)^p} = \frac 1{(1-p) (\ln 2)^p}$ for all $\displaystyle p \ne 1$.

So the entire integral comes to $\displaystyle -\frac 1{(1-p) (\ln 2)^p}$, which means that the series converges when that fraction converges. Right so far? If so, now what?