Determining the convergence or divergence of a sequence using direct comparison

[tmt1]

I have

$$\sum_{n = 2}^{\infty} \frac{{(\ln\left({n}\right)})^{12}}{n^{\frac{9}{8}}}$$

We can compare it to $ \frac{1}{{n}^{\frac{1}{8}}}$. $ \sum_{n = 1}^{\infty} \frac{1}{{n}^{\frac{1}{8}}}$ diverges because $p < 1$ in this case. So, if I can prove that $ \frac{{(\ln\left({n}\right)})^{12}}{n^{\frac{9}{8}}} \ge \frac{1}{{n}^{\frac{1}{8}}}$ then that would mean $\sum_{n = 2}^{\infty} \frac{{\ln\left({n}\right)}^{12}}{n^{\frac{9}{8}}}$ diverges. Or $ \frac{{(\ln\left({n}\right)})^{12}}{n^{\frac{9}{8}}} - \frac{1}{{n}^{\frac{1}{8}}} \ge 0 $.

How can I prove this?

 

[Opalg]

I have

$$\sum_{n = 2}^{\infty} \frac{{(\ln\left({n}\right)})^{12}}{n^{\frac{9}{8}}}$$

We can compare it to $ \frac{1}{{n}^{\frac{1}{8}}}$. $ \sum_{n = 1}^{\infty} \frac{1}{{n}^{\frac{1}{8}}}$ diverges because $p < 1$ in this case. So, if I can prove that $ \frac{{(\ln\left({n}\right)})^{12}}{n^{\frac{9}{8}}} \ge \frac{1}{{n}^{\frac{1}{8}}}$ then that would mean $\sum_{n = 2}^{\infty} \frac{{\ln\left({n}\right)}^{12}}{n^{\frac{9}{8}}}$ diverges. Or $ \frac{{(\ln\left({n}\right)})^{12}}{n^{\frac{9}{8}}} - \frac{1}{{n}^{\frac{1}{8}}} \ge 0 $.

How can I prove this?

The guiding principle is that $\ln n$ increases more slowly than any (positive) power of $n$. So for convergence purposes $\sum\frac{(\ln n)^{12}}{n^{9/8}}$ behaves pretty much like $\sum\frac1{n^{9/8}}$. Can you take it from there?