Convergence of $\sum_{n=2}^\infty \frac {1}{(\log n )^3}$ Test Convergence of $\displaystyle\sum_{n=2}^\infty \dfrac {1}{(\log n )^3}$
Attempt: I haven't been able to find a suitable comparator for $\dfrac {1}{(\log n )^3}$ . The integration test also seems tedious.
Please guide me on how to move forward. 
Thank you very much for your help.
 A: HINT: $\log x<x^{1/4}$ for large $x$.
A: Notice that $f(n):=\frac{1}{\log^{3}(n)}>0$ $\forall n\geq2$ so we will be able to use the comparison test. Furthermore, $\exists c\in[2,\infty)$ such that $n>\log^{3}(n)$ $\forall n>c$. Hence $\frac{1}{n}<\frac{1}{\log^{3}(n)}$ $\forall n>c$ and as we know that $\sum\limits_{n=c}^{\infty}\frac{1}{n}$ is divergent, by the comparison test $\sum_\limits{n=c}^{\infty}f(n)$ must diverge too. Then, $\sum_\limits{n=2}^{\infty}f(n)$ diverges.
A: In general, we have this $\displaystyle\sum_{n=2}^\infty \dfrac {1}{(\ln n )^k}$, where  $k\in\Bbb Z.$ To see this, we know $\ln (x^a)=a\ln(x)$ and $\ln (x^a)<x^a$. So, $a \ln(x)<x^a.$ Then, $$\frac{1}{\ln x}>\frac{a}{x^a}$$, put $x=n^k$. So, $$\frac{1}{\ln^k n}>\frac{a}{n^{ka}},$$ since $a$ is arbitrary, let $a=\frac{1}{k}$. We have  $$\frac{1}{\ln^k n}>\frac{1}{k n},$$ by using the comparsiion test
$\displaystyle\sum_{n=2}^\infty \dfrac {1}{(k n)}$ is divergent. Then $\displaystyle\sum_{n=2}^\infty \dfrac {1}{(\ln n )^k}$ is divergent as needed.
