# Math Help - Series: Converge of Diverge

1. ## Series: Converge of Diverge

Determine whether the series converges or diverges.

Summation of (ln(k))/(k^3)

2. Originally Posted by WartonMorton
Determine whether the series converges or diverges.

Summation of (ln(k))/(k^3)
Since $|\ln(x)| for $x>1$ the general general term is positive and $<1/k^2$ and you should be able to take it from there

CB

3. Originally Posted by WartonMorton
Determine whether the series converges or diverges.

Summation of (ln(k))/(k^3)

hint: i guess this should work (ln(k))/(k^3) < k/(k^3) = 1/k^2

4. Originally Posted by WartonMorton
Determine whether the series converges or diverges.

Summation of (ln(k))/(k^3)
Oh sure it converges. This is because $\frac{\ln k}{k^3}=\frac{\ln k}{k}\cdot\frac{1}{k^2}$, where the first factor $\frac{\ln k}{k}$ converges to $0$ for $k\to\infty$ and the series $\sum_k \frac{1}{k^2}$ of the second factor $\frac{1}{k^2}$ also converges. - But maybe this way of arguing for convergence is a little too sloppy for your taste.