explore the convergence of series with ln(n)

Question

Help me explore the convergence of red color rounded series. On this photo (the equation below) I used radical indication but it doesn't show me the result. What would be better to use?

$$\color{red}{\sum_{n=1}^{\infty}\frac{1}{n^3\ln(n)}=\text{?}}$$ $$\lim_{n \to \infty}\sqrt[n]{\frac{1}{n^3\ln(n)}} = \lim_{n \to \infty}\frac{\sqrt[n]{1^n}}{\sqrt[n]{3^n} \sqrt[n]{\ln(n)}} = 1$$

Radical Cauchy indication does not help.

Answer 1

Use the comparison with the Riemann series $$\frac1{n^3\ln n}\le \frac1{n^3}\quad\text{for $n$ large enough}$$

Answer 2

Since $\log (n) >1$ as soon as $n \geq 3$, you can write $$\sum _{n=2}^{\infty } \frac{1}{n^3 \log (n)}=\frac{1}{8 \log (2)}+\sum _{n=3}^{\infty } \frac{1}{n^3 \log (n)}$$ and now use the comparison suggested by Sami Ben Romdhane.

Added after Raymond Manzoni's comments

Just for entertainment, $$\sum _{n=3}^{\infty } \frac{1}{n^3}=\zeta (3)-\frac{9}{8}=0.077057$$ So,$$\sum _{n=2}^{\infty } \frac{1}{n^3 \log (n)}=0.237996 \le \frac{1}{8 \log (2)}+\zeta (3)-\frac{9}{8}=0.257394$$