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 $\displaystyle\sum\limits_{n=4}^{\infty }\dfrac{\sqrt{n}}{n-3}$ ?**
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