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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.

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2 Answers 2

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Nutshell: ultimately what matters is that you show $(lnx)^3<x$ 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?

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  • $\begingroup$ 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 $\ln(x)^2$. $\endgroup$
    – zwim
    Commented May 1, 2020 at 21:16
  • $\begingroup$ @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 $\endgroup$
    – imranfat
    Commented May 1, 2020 at 21:37
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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 $n \ge 94$.

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