# Series converge or diverge

zeion

## Homework Statement

Determine whether the series converges or diverges.

$$\sum \frac{lnk}{k^3}$$

## The Attempt at a Solution

Since lnk always less than 0, so $$\frac{lnk}{k^3} \leq \frac{1}{k^3}$$ and $$\frac{1}{k^3}$$diverges so$$\frac{lnk}{k^3}$$ diverges.

Mentor

## Homework Statement

Determine whether the series converges or diverges.

$$\sum \frac{lnk}{k^3}$$

## The Attempt at a Solution

Since lnk always less than 0, so $$\frac{lnk}{k^3} \leq \frac{1}{k^3}$$ and $$\frac{1}{k^3}$$diverges so$$\frac{lnk}{k^3}$$ diverges.
I count three mistakes here.
1. Why do you think that ln(k) is always < 0? I'm assuming that k = 1, 2, 3, ...
2. The series whose general term is 1/k3 converges.
3. If you want to show that a given series diverges, its terms must be larger than those of a divergent series. If the terms of the given series are smaller than those of a divergent series, you can't conclude anything.

zeion
Can I argue that $$ln(k) < k$$ then $$\frac{ln(k)}{k^3} < \frac{k}{k^3}$$then it converge because $$\frac{1}{k^2}$$ converges?

Mentor
I can buy that. However, your should be able to convince yourself that ln(x) < x.

zeion
Yeah since ln(x) is the inverse of e^x and they reflect on the y =x.

Mentor
The fact that y = ln(x) and y = ex are reflections across the line y = x doesn't prove that ln(x) < x. If y = ex happens to cross this line a few times means that ln(x) will do so, also.