The series $\sum_{n=2}^\infty 1/(logn)^p$

Question

$\sum_{n=2}^\infty 1/(logn)^p$ is similar to $\sum_{n=2}^\infty 1/n^p$

Then is it convergent when p>1 ? like p-test?

Answer 1

Using the l'Hospital rule

$$\lim_{n\to\infty}\frac{\ln^p n}{n}=0$$ so for sufficiently large $n$ we have

$$\frac{1}{\ln^pn}\ge \frac1n$$ and we can use comparison with the harmonic series to conclude.

Answer 2

No, it's never convergent. Note that, since $$ \log\frac{n}{\log^p n} = \log n - p\log\log n \to \infty $$ we have $$ \frac{n}{\log^p n} \to \infty $$ and hence there is $N$ $$ \frac 1n \le \frac 1{\log^p n}, \quad n \ge N $$ As $\sum \frac 1n$ diverges, $\sum \log^{-p} n$ also does.