Convergence/divergence of series $\sum\limits_n\frac1{(\log(1+n))^p}$

Question

$$\sum u_n\qquad u_n = \frac{1}{(\log(1+ n))^p}$$ Wolfram says it's convergent, my book proves it divergent, someone said it depends on p. Can anyone help with the actual answer?

Answer 1

It is obviously divergent: $(\log(n))^p<n$, for any $p>0$ and great enough $n$. On the other hand, if $p\leq 0$, the general term increases, thus the series diverges.

Answer 2

We have that for $p\le 0$

$$\frac{1}{(\log(1+ n))^p}\not \to 0$$

and the series doesn't converge.

For $p>0$, as an alternative, by Cauchy condensation test since

$$ 2^na_{2^n}=\frac{2^n}{(\log(1+ 2^n))^p} \sim \frac{2^n}{n^p(\log 2)^p} \to \infty$$

the condensed series $\sum 2^na_{2^n}$ diverges and then also the given series diverges.