## Calculus 8th Edition

Converges to $-2$
Given:$a_n=n-\sqrt {n+1}\sqrt {n+3}$ $a_n=n-\sqrt {(n+1)(n+3)}$ $a_n=n-\sqrt {n^{2}+4n+3}$ Multiply numerator and denominator by the conjugate. $n-\sqrt {n^{2}+4n+3}\times \frac{n+\sqrt {n^{2}+4n+3} }{n+\sqrt {n^{2}+4n+3}}=\frac{-4n-3}{n+\sqrt {n^{2}+4n+3}}$ $=\frac{-4-\frac{3}{n}}{1+\sqrt {1++\frac{4}{n}+\frac{3}{n}}}$ Find the limit as $n \to \infty$. $\lim\limits_{n \to \infty}\frac{-4-\frac{3}{n}}{1+\sqrt {1++\frac{4}{n}+\frac{3}{n}}}=\frac{-4-0}{1+\sqrt {1+0+0}}$ $=\frac{-4}{2}$ $=-2$ Hence, the sequence converges to $-2$.