Numerical Series nature I need someone to approve if my way to prove that the following numerical series is divergent. 


*

*Consider the Series $\sum_{n=1} \frac{1}{ln(n+n^2+n^3)}$. following the comparison test i found that the series diverge.

*My approach:
since  $e^n +\frac{n^2}{2!} +\frac{5n^3}{3!} \ge n+n^2+n^3$ then    $ \ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!}) \ge ln(n+n^2+n^3) $ 
i.e $\frac{1}{ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!})} \le \frac{1}{ln(n+n^2+n^3)}$ but since $e^n +\frac{n^2}{2!} +\frac{5n^3}{3!} \sim e^n$  as $n\rightarrow \infty$ 
hence$\frac{1}{ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!})} \sim \frac{1}{n}$ where $\sum\frac{1}{n}$ is divergent according to Reimann and so we get that the sum $\sum_{n=1} \frac{1}{ln(n+n^2+n^3)}$ is also divergent. 
Am i right doing so ?

$\log(n+n^2+n^3)\leq\log(n^3+n^3+n^3)=\log(3n^3)\leq\log(n^4)=4\log(n)\leq 4n$ if $n\geq 3$. Hence
$$
\sum_{n=1}^{\infty}\frac{1}{\log(n+n^2+n^3)}\geq\frac{1}{4}\sum_{n=3}^{\infty}\frac{1}{n}
$$
which is divergent.
