Question

Determine if the following infinite series is convergent or divergent:

Answer 1

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

Answer 2

hint:ln x is less than sqrt x

Answer 3

use integral test
hint use u=ln n

Answer 4

is it \(\frac{1}{\ln(n)\ln(\ln(n))}\)?

Answer 5

\[(\ln)^\ln(\ln n)=e^\ln(\ln n)^\ln(\ln n)=e^[\ln(\ln n)]^2\]

Answer 6

and by the hint ln(ln n) is less than sqrt(ln n) implies....1/e[ln(ln n)]squared is greater than 1/e^ln n all equals to 1/n

Answer 7

i can't figure out what it is you are writing, but it diverges for sure, because the log grows very slowly

Answer 8

it diverges and im sorry for the dashes in my work i meant to put n in those so sorry for the sloppiness

Answer 9

maybe
\[\frac{1}{\ln(n)^{\ln(ln(n))}}\]?

Answer 10

correct

Answer 11

it is converegent
because