We know that
$0<\dfrac{n^2-2}{n^6+1}<\dfrac{n^2-2}{n^6}<\dfrac{n^2}{n^6}=\dfrac{1}{n^4}$
for any $n\ge 2$.
**Considering this fact, what does the direct comparison test say about $\displaystyle\sum_{n=2}^{\infty }{\frac{n^2-2}{{{n}^{6}}+1}}$ ?**
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