Infinite series with natural log to a high power divided by an exponential

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.

• 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. Commented Nov 8, 2023 at 21:55
• 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. Commented Nov 8, 2023 at 22:20

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 .

\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.