Question

In the following series, what values of p make it converge?

Answer 1

\[\sum_{n=2}^{\infty} \frac{ 1 }{ n ^{p} \ln(n) }\]

Answer 2

Let a_n = 1/ n^p * ln(n) and b_n = 1/ n^p. Note that a_n and b_n are always positive for n>= 2.

Answer 3

n^p * ln(n) > n^p so..
1 / n^p * ln(n) < 1/ n^p
So the comparison test applies.

Answer 4

Can I just use the p-series to say that it is convergent if p>1 and divergent if p<=1 ? But the p-series starts at n=1 instead of n=2 so I don't know if that would make a difference..

Answer 5

Have you considered using the integral test?

Answer 6

I think there's a loophole somewhere... It seems to easy to be true... OR maybe instead I can use the integral test on the 1/ n^p?

Answer 7

Yeah that's what I was thinking.. it seems to be a safer route.

Answer 8

But to use the integral test 1/n^p has to be continuous... Not sure how to show that...

Answer 9

It is for \(n\ge2\). The discontinuity is at \(n=0\), but that's not a term in your series.

Answer 10

Ohh of course.

Answer 11

So would we start our integral on [1, inf)?

Answer 12

It's part of the Integral Test I think?

Answer 13

The integral should start wherever the series starts. If \(n=1\), you have \(0\) in the denominator. We'll want to examine
\[\int_2^\infty\frac{dn}{n^p\ln n}\]

Answer 14

Okay so ditch the comparison theorem and instead just do the integral test for the given function in the series?

Answer 15

OH and I understand why it doesn't always start at n=1. Thanks (:

Answer 16

More generally, the integral test can be used for any continuous, positive, decreasing function \(f(n)\) on the interval \([k,\infty)\) to make a conclusion for the series \(\sum\limits_{n=k}^\infty a_n\)

Answer 17

If the integral test works for a series starting at \(n=1\), it should also work for a series starting at any value of \(n>1\).
\[\sum_{n=1}^\infty a_n=a_1+\sum_{n=2}^\infty=a_1+a_2+\sum_{n=3}^\infty a_n=\cdots\]
That's not to say the comparison test doesn't work. You just have to be careful about what you say. For example, \(\dfrac{1}{n^p\ln n}\) is not smaller than \(\dfrac{1}{n^p}\) if \(n=2\) because \(\ln2<1\).

Answer 18

I see, so I'll try doing the integral test then rn.