# Comparing the harmonic series with $\sum_{n=2}^{\infty} \frac{1}{{(\ln n)}^{3}}$

I have been asked at what point are the terms of the harmonic series less than the terms of the following series:

$$\sum_{n=2}^{\infty} \frac{1}{{(\ln n)}^{3}}$$

I understand that this is to prove that the above series diverges, which I can do, but I am confused about how to get the exact value of n, where the above series would become greater than the harmonic series.

Nutshell: ultimately what matters is that you show $$(lnx)^3 for some $$x$$ which is not so difficult. Bring the $$x$$ to the left and take a derivative and show that there is an $$x$$ after which the derivative is negative. Now you have a comparison because then you know that there is an $$x$$ for which $$1/x$$ is greater than $$\frac{1}{(lnx)^3}$$. Using comparison, now add detail and can you finish the problem?
• I do not agree with the "not so difficult". It is in fact easier to study $x^{1/3}-\ln(x)$ than $x-\ln(x)^3$. In the first case sign of derivative it easy to determine (max in $x=27$) while in the second case you are stuck with comparing a similar problem between $x$ and $\ln(x)^2$.
• @zwim. Of course, you have a point. But it is not necessary that an exact value for x is obtained. In fact, all that matters is that there exists an x for which the derivative becomes negative afterwards. The OP stated that he wants to find an exact value of $n$. Not sure why, as this is not relevant for proving convergence/divergence Commented May 1, 2020 at 21:37
Your calculator will tell you $$93 < \ln(93)^3$$ while $$94 > \ln(94)^3$$, and since $$x - \ln(x)^3$$ is convex for $$x > e^2$$ you will have $$n > \ln(n)^3$$ for all $$n \ge 94$$.