We know that $\displaystyle\frac{1}{\sqrt{n}-5}>\frac{1}{\sqrt{n}}>0$ for any $n\ge 26$. **Considering this fact, what does the direct comparison test say about $\displaystyle\sum\limits_{n=26}^{\infty }{\frac{1}{\sqrt{n}-5}}$ ?** [[☃ radio 3]]