# Series: Converge of Diverge

• May 6th 2010, 03:59 AM
WartonMorton
Series: Converge of Diverge
Determine whether the series converges or diverges.

Summation of (ln(k))/(k^3)
• May 6th 2010, 04:02 AM
CaptainBlack
Quote:

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

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

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

CB
• May 6th 2010, 04:05 AM
aman_cc
Quote:

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
• May 6th 2010, 04:05 AM
Failure
Quote:

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

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

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