How do I prove that the sum: $1/\ln(n)^p$ diverges for $p>1$ So I need to prove that the infinite sum $\frac{1}{(\ln(n)^p)}$ diverges for all values of $p$. I managed to prove it for $p\leq 1$ via comparison test with $1/n$. but for this I can't seem to find a way to prove it diverges for $p>1$.
 A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 <x\tag 1$$
for $0<x$.  

Using the property $\log(x^a)=a\log(x)$ in $(1)$, we obtain for $a>0$
$$\log(x)<\frac{x^a}{a}\tag2 $$
For $x>n>1$, we have from $(2)$
$$\frac{1}{\log^p(n)}>\frac{a}{n^{ap}}\tag 1$$
For any $p>1$, we can take $a=1/p$ so that $ap=1$.  Hence, the series $\sum_{n=1}\frac{1}{\log^p(n)}$ dominates the series, $\frac1p\sum_{n=2}^\infty\frac{1}{n}$, which diverges.  The comparison test guarantees that the series $\sum_{n=1}\frac{1}{\log^p(n)}$ diverges also.
A: Use that for $n$ large enough $\ln^p n < n$ and compare your series with $\sum 1/n$.
A: You can prove it using the following theorem

Theorem (the integral test): Let $f(x)$ be a continuous function of real numbers that is monotonic decreasing. Then for any $N\in\mathbb N$, $\sum_{n=N}^\infty f(n)$ converges if and only if $\int_N^\infty f(x)dx$ is finite.

