We know that $\displaystyle\dfrac{\sqrt{n}}{n-3}>\dfrac{\sqrt{n}}{n}=\dfrac{1}{\sqrt{n}}>0$ for any $n\ge 4$. **Considering this fact, what does the direct comparison test say about $\displaystyle\sum\limits_{n=4}^{\infty }\dfrac{\sqrt{n}}{n-3}$ ?** [[☃ radio 3]]