1. determine if it's converge

$\displaystyle \sum (ln k)^2/k$

use the root test or the ration test

2. Try the integral test

3. Originally Posted by wopashui
$\displaystyle \sum (ln k)^2/k$

use the root test or the ration test
see attachment for solution:series converges

4. Originally Posted by wopashui
$\displaystyle \sum (ln k)^2/k$

use the root test or the ration test
What values does $\displaystyle k$ go between?

Is it $\displaystyle \sum_{k = 1}^{\infty}\frac{(\ln{k})^2}{k}$?

5. Originally Posted by wopashui
$\displaystyle \sum (ln k)^2/k$

use the root test or the ration test
Before using blindly any "test", simply look at what the summand looks like. It is positive. The logarithm grows slowly to $\displaystyle \infty$, hence $\displaystyle \frac{(\ln k)^2}{k}$ is a bit larger than $\displaystyle \frac{1}{k}$. But... wait! we know that $\displaystyle \sum_k \frac{1}{k}$ diverges! So we just have to write $\displaystyle \frac{(\ln k)^2}{k}\geq \frac{1}{k}$ when $\displaystyle k\geq 3$ to conclude: the series diverges.