Question

For what values of k does the following series absolutely converge and for what values of k does the series conditionally converge?

Answer 1

\[\sum_{n=3}^{\infty} \frac{ (-1)^n }{ n*\ln(n)*(\ln(\ln(n)))^k }\]

Answer 2

@satellite73 any advice would be greatly appreciated

Answer 3

wow

Answer 4

Conditional convergence values of k for k>0

Answer 5

it certainly will converge conditionally if \(k>0\) because it alternates and the terms go to zero

Answer 6

oh you said that

Answer 7

so the real question is, for what k does \[\sum_{n=3}^{\infty} \frac{ 1 }{ n*\ln(n)*(\ln(\ln(n)))^k }\]converge

Answer 8

i can make a guess, but that is no help

Answer 9

If the sum of the absolute value converges then those values of k indicate it is absolutely convergent right?

Answer 10

yes, i.e. get rid of the \((-1)^n\) and replace it by \(1\)

Answer 11

fraid i am not a lot of use for this one
my guess is \(k>1\) but it is just a guess

Answer 12

you got a test to use?

Answer 13

ooh how about the integral test?

Answer 14

using a u sub twice for \[\int \frac{1}{x\log(x)\log(\log(x))}dx\] you get
\[\log(\log(\log(x)))\]

Answer 15

so diverges at \(k=1\) but probably get something nicer at \(k=1+\epsilon\)

Answer 16

yeah i think the log ends up in the denominator when you do that

Answer 17

try it, i bet it works

Answer 18

Yup after the second substitution it comes to ln(ln(x)^1-k / 1-k

Answer 19

whew, thought it was going to be a complete stumper