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 $\displaystyle\sum_{n=2}^{\infty }{\frac{n^2-2}{{{n}^{6}}+1}}$ ?** [[☃ radio 3]]