How do I show that the series $\sum_{n=2}^\infty \frac{1}{n(\log n)^p}$ is convergent or divergent?

Question

How do I show that the series $\sum_{n=2}^\infty \frac{1}{n(\log n)^p}$ is convergent or divergent?

I've done it for $p \ge 0$ using condensation test but I can't proceed for $p \lt 0$.

Any suggestions?

Answer 1

If $p<0$, then $(\log n)^p<1$, and therefore$$\frac1{n(\log n)^p}>\frac1n.$$Now, use the fact that the harmonic series diverges.

Answer 2

When $p < 0$ the summand will (eventually) be greater than $1/n$, so the series will diverge.

Answer 3

If $p<0$, then $$\frac 1{n(\log n)^p}>\frac 1n$$ Since $\sum \frac 1n$ diverges, so must $\sum \frac 1{n(\log n)^p}$