# 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?

• I think it converges, by direct application of the integral test.
– lulu
Apr 2 '18 at 13:23
• Note: I assumed that $\ln^3(x)$ meant $\left(\ln(x) \right)^3$. Perhaps you meant something else? In any case, you should edit to clarify.
– lulu
Apr 2 '18 at 13:32
• @lulu This exercise comes from a book. I have noticed the author to write ($\ln\ln(x)$) when he wants the composition, so I assume here that the author is referring to $\ln^3(x)=(\ln(x))^3$ as you stated. I am little bit confused regarding the solution once it states that the series converge. Apr 2 '18 at 13:37
• Well, the posted solutions seem comprehensive. If you understand it to mean the cube you get a convergent series, if you understand it to mean composition you get a divergent series.
– lulu
Apr 2 '18 at 13:39
• If $\ln^3 x$ means $\ln(\ln(\ln x)$ then the series is divergent. Apr 2 '18 at 13:41

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).

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.

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 $f(n)$: $$\sum_{n=0}^{\infty} f(n) \hbox{ converges iff } \sum_{n=0}^{\infty} 2^n f(2^n) \hbox{ converges }$$

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.