We know that
$\displaystyle \dfrac{n^2}{n^3-\ln(n)} \ge \dfrac{n^2}{n^3}=\dfrac{1}{n}>0$
for any $n\ge 1$.
**Considering this fact, what does the direct comparison test say about $\displaystyle\sum\limits_{n=1}^{\infty }~\dfrac{n^2}{n^3-\ln(n)}$ ?**
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