## Calculus (3rd Edition)

Given $$\sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{2}-3}}$$ We compare the given series with the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n }$, a divergent $p-$series with $p=1$ and for $n\geq 2$ $$\frac{1}{\sqrt{n^{2}-3}}\geq\frac{1}{n}$$ Then $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{2}-3}}$ also diverges.