<s>[INST] 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 ?
 [/INST] $$\log{(n+n^2+n^3)} = \log{n^3} + \log{\left (1+\frac1{n}+\frac1{n^2} \right )} = 3 \log{n}+O\left ( \frac1{n} \right )$$
Thus
$$\frac1{\log{(n+n^2+n^3)}} = \frac1{3 \log{n}} + O\left ( \frac1{n \log{n}} \right )$$
Thus the sum diverges by comparison with, e.g., then sum over $1/n$ which diverges.
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