Question

I'm looking for some clues on what Series Convergence Test to use for this series, so far I have eliminated the Integral Test and the Limit Comparison test.

Answer 1

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

Answer 2

Have you tried the ratio test? Just a thought.

Answer 3

I will try that quickly.

Answer 4

Never mind, the ratio test fails.

Answer 5

Could you help me try the Comparison test?

Answer 6

Yes, I was just about to suggest that. Try comparing it to
\[\sum_{n=2}^\infty\frac{1}{\sqrt n}\]

Answer 7

That's actually the algebraic result i got for the Limit Comparison Test but ill try the comparison test

Answer 8

Does that count as a p-series?

Answer 9

Yes it does. p = 1/2, in the case of the series I suggested, so it diverges.

Answer 10

You have to show that
\[\sum_{n=2}^\infty\frac{1}{\sqrt n}<\sum_{n=2}^\infty\frac{1}{\sqrt n \ln(n)}\]

Answer 11

That does not appear to be true

Answer 12

Oh, my mistake. I had the inequality wrong in my mind. I'll try to see what else can be done.

Answer 13

I might be wrong as well, I just took an arbitrary number and plugged it into both and it looks like 1/sqrt(n) comes out greater

Answer 14

no way this converges, by eyeball test

Answer 15

but i suppose that is not much help
maybe you can use \(\ln(n)<\sqrt{n}\) and so
\[\frac{1}{\sqrt{n}\ln(n)}>\frac{1}{n}\] so diverges via comparison test