calculus - Prove $sumlimits_{n=2}^infty frac {1} {nln(n)sqrt{ln^3{n}}}$ 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!

Answer

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.