Prove $\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\sqrt{\ln^3{n}}}$ is divergent. 
Evaluate if the following series is convergent or divergent: $\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\sqrt{\ln^3{n}}}$.

After checking the solution I found out the series was divergent. I tried to use the comparison test or Weierstrass's test to evaluate the series. I started by using the inequality $\ln(n)\leqslant n$ in the following way:
$\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\sqrt{\ln^3{n}}}>\sum\limits_{n=2}^\infty \frac {1} {n^2+\sqrt{n^3}}>\sum\limits_{n=2}^\infty \frac {1} {n^2+{n^3}}$.
Since $\frac {1} {x^2+{x^3}}$ is monotone decreasing I computed: $\int_\limits{1}^{\infty}\frac {1} {x^2+{x^3}}=\int_\limits{1}^{\infty}-\frac {1} {x}+\frac{1}{x+1}+\frac{1}{x^2}=1-\ln(2)$, so the series $\sum\limits_{n=2}^\infty \frac {1} {n^2+{n^3}}$ converges. I tried to find a series in between that would diverge but I have not come to an idea of what the numerator should be.
Question:
How can I prove the series $\sum\limits_{n=2}^\infty \frac {1} {n\ln(n)\sqrt{\ln^3{n}}}$ to be divergent?
Thanks in advance!
 A: It is well-known a Bertrand's series:
$$\sum_{n\ge 2}\frac 1{n^\alpha\log^\beta\! n}$$
converges if and only if


*

*$\alpha>1$ (by comparison with the Riemann series $\;\sum_{n}\frac 1{n^\alpha}$);

*or $\alpha=1$ and $\beta>1$ (by the integral test).

A: For series $\dfrac 1{n^\alpha\ln(n)^\beta}$ the easiest test is the Cauchy condensation test. 
In this case for $\alpha=1$ and $\beta=\frac 52>1$ it should converge. 
Unless your $\ln^3 n$ meant $\ln(\ln(\ln(n)))$ in which case you would have $\dfrac 1{n^\alpha\ln(n)^\beta\ln(\ln(\ln(n)))^\gamma}$ with $\alpha=\beta=1$ and $\gamma=\frac 12<1$ so it is divergent.
A: Does $ln^3(x)$ mean iterated 3 times or cubed? If cubed I think is false, if iterated 3 times true.
Hint: (Cauchy condensation test.) For a non-negative, decreasing sequence of reals $f(n)$:
$$\sum_{n=0}^{\infty} f(n) \hbox{    converges iff } \sum_{n=0}^{\infty} 2^n f(2^n) \hbox{     converges }$$
A: Use Cauchy's condensation test.
$$\frac {2^n} {2^n\ln(2^n)\sqrt{\ln^3{2^n}}} = \frac{1}{n\ln2 \cdot (n \ln2)^{3/2}} = \frac{1}{(\ln2)^{5/2} \, n^{5/2}}$$
So it's convergent.
