What value of p does the following series converge

**MSUMathStdnt** · December 19th 2012, 04:27 PM

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

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

The answer is $p=1$ (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 $\frac 1{n (\ln n)^p}$ to $\frac {1}{n^p}$ . If $\frac 1{(n \ln n)^p} < \frac {1}{n^p}$ , then the former converges, too. For $p>1$ , and since $n>1$ , then $(\ln n)^p>1$ and thus $\frac 1{(n \ln n)^p} < \frac {1}{n^p}$ . How am I doing so far?

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

**chiro** · December 19th 2012, 05:07 PM

Hey MSUMathStdnt.

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

**Plato** · December 19th 2012, 05:09 PM

Originally Posted by MSUMathStdnt

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

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

The answer is $p=1$ (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.

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

Look at the integral test.

So what is your question now?

**MSUMathStdnt** · December 19th 2012, 05:28 PM

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

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

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

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

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

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

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