## Calculus (3rd Edition)

Given $$\sum_{n=4}^{\infty}\frac{\sqrt{n} }{n-3}$$ Compare with $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}$, a divergent series ( $p-$series with $p<1$) and for $n\geq 4$ \begin{align*} \frac{\sqrt{n} }{n-3}&\geq \frac{1 }{\sqrt{n}} \end{align*} Then $\displaystyle\sum_{n=4}^{\infty} \frac{\sqrt{n} }{n-3}$ also diverges.