Question

Determine whether the series Converges or Diverges

Answer 1

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

Answer 2

Am I correct in saying that I have to use the integral test?

Answer 3

I wonder if I could use the comparison test though...

Answer 4

Shall I compare with the harmonic series?

Answer 5

Because it diverges.

Answer 6

I think the integral test works. Since since n ln n is monoticonally increasing, 1/(n ln n) is monotonitcally decreasing. Also, it is non-negative. But I think it is more work.

Yeah I would suggest the comparison test.

Answer 7

Yeah. So am I correct in saying that I should compare this to the harmonic series? It's pretty similar.

Answer 8

Because for large values of n, 1/n dominates because ln(n) will be small compared to 1/n.

Answer 9

Hm wait I'm not sure if the comparison with 1/n works. Because n ln n > n for n > e
1/ (n ln n) < 1/n and 1/n diverges, but it's the larger series so you can't conclude that the smaller series diverges

Answer 10

What comparison would you suggest then?

Answer 11

I'm trying to think of something. But you know what The integral test might not be so bad here because substituting u = ln then du = 1/n which is in your integral. It's actually not that bad of an integration.

Answer 12

Hmm okay...

Answer 13

Ahh I got it. :) .

Answer 14

I end up getting:

\[\lim_{t \rightarrow \infty}(\frac{ \ln ^2(t)-\ln ^2(t) }{ 2 })\]

Answer 15

Which clearly diverges.

Answer 16

Yep :)

Answer 17

I'm not sure if there's an easy comparison to use on this one. I tried googling and found a website using comparison tests which uses this exact series to compare other series, but they used the integral test on this series to prove its divergence... lol

Answer 18

lol. Wolfram uses the comparison test so I tried to find one.

Answer 19

Ah.. wolfram :) lol. I wish I knew what it used