Question

sum from n=2 to infinity (1/(n(ln)^2))

Answer 1

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

Answer 2

sorry n=2

Answer 3

So what's the question?

Answer 4

what i get is 1/(ln n) for the integral

Answer 5

doesn't seem right

Answer 6

Ah, so it's a question of convergence or divergence. Right?

Answer 7

correct I didn't plug the numbers in yet because I believe i didn't do the integral right

Answer 8

\[\int_2^\infty \frac{dx}{x(\ln x)^2}=\int_{\ln 2}^\infty \frac{du}{u^2}=\left.\frac{-1}{u}\right |_{\ln 2}^\infty=\ln 2\]
You actually did the integral correctly.

Answer 9

but I got (1/ ln x)

Answer 10

Then you evaluate that between 2 and infinity and you get the same answer that I did.

Answer 11

Thanks!

Answer 12

I believe you made an error though, I still get - (1/ ln (2)) after evaluating the integral from 2 to infinity

Answer 13

Well it's convergent in the end...