I am asked to prove the divergence or convergence of this series using any of these methods: Telescoping, Geometric Series, Divergence Test, Integral Test, Direct Comparison, and Limit Comparison. The series is: $\sum_{n=2}^\infty \frac{(\ln(n))^{12}}{n^{\frac{9}{8}}}$

I can say I have eliminated the possibility of telescoping and geometric series. I have performed the divergence test and the limit of the sequence approaches 0, so I know that I need to do more test to actually determine if it diverges or converges. I have attempted the integral test, however, I am not really sure if it is possible to integrate the function (maybe a really drawn out integration by parts? I haven't tried because I believe there has to be a cleaner method). I have tried the direct comparison but struggle to choose a good comparison for the test, my initial thought was comparing to an equation with a denominator raised to a power slightly lower, but still greater than 1 so it would converge via p-test. That does not work due to the variable in the numerator. On the limit comparison test, all my comparisons lead to inconclusive results on the behavior of the series. I am sure there is probably an easy or interesting solution, but I need help to find it.

  • $\begingroup$ I'm sorry about the formatting I actually don't know how to get it the way I want it to display, I am just hoping that someone can just revise that. I tried to use the MathJax tutorial. $\endgroup$
    – NotVersed
    Commented Nov 8, 2023 at 21:55
  • 1
    $\begingroup$ I have attempted to fix your MathJax, please check if it's the correct expression. You can click the edit link to see how it is formatted. $\endgroup$
    – ConMan
    Commented Nov 8, 2023 at 22:20

2 Answers 2


You can (and should) show that $\ln n$ grows slower than $n^k$ for any $k > 0$, i.e. $\lim_{n \rightarrow \infty} \frac{\ln n}{n^k} = 0$.

Based on that, you should then be able to choose a suitable value of $k$ so that you can use the Ratio Test comparing $\frac{(\ln n)^{12}}{n^\frac{9}{8}}$ to $\frac{1}{n^k}$.


$$\sum_{n=2}^\infty \frac{(\ln(n))^{12}}{n^{\frac{9}{8}}}$$ is the first point: here we need to find out if the series converges, diverges, by using the Direct Comparison Test.

We choose $N$ with $n \ge N$ thus $\ln n <n^{\frac{1}{192}}$.

\begin{align*} (\ln n)^{12}<&\, (n^{\frac{1}{192}})^{12}\\ (\ln n)^{12}<&\,n^{\frac{1}{16}} \end{align*}

$$\sum_{n=2}^\infty \frac{(\ln(n))^{12}}{n^{\frac{9}{8}}}\leq \sum_{n=2}^\infty \frac{n^{\frac{1}{16}}}{n^{\frac{9}{8}}}=\sum_{n=2}^\infty \frac{1}{n^{\frac{17}{16}}}$$

Convergence $p$-series: the infinite series $$\sum_{n=1}^\infty \frac{1}{n^p}$$ converges if $p>1$ and if $p ≤ 1$, then the series diverges.

Being here $p=\frac{17}{16}>1$, we can conclude that the series is convergent.


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