We know that $\displaystyle \dfrac{2}{2n^5-3} > \dfrac{2}{2n^5} = \dfrac{1}{n^5}>0$ for any $n\ge 2$. **Considering this fact, what does the direct comparison test say about $\displaystyle\sum\limits_{n=2}^{\infty }\dfrac{2}{2n^5-3}$ ?** [[☃ radio 3]]