Answer
Convergese to $-2$
Work Step by Step
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 $-2$.
