Answer
The general term is ${a_n} = \left( {\sqrt {n + 1} - \sqrt {n + 2} } \right)$.
The sequence converges.
The limit of the sequence:
$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {n + 1} - \sqrt {n + 2} } \right) = 0$
Work Step by Step
1. Starting with $n=1$, the general term is
${a_n} = \left( {\sqrt {n + 1} - \sqrt {n + 2} } \right)$
2. Replacing $n$ by $x$, write $f\left( x \right) = \left( {\sqrt {x + 1} - \sqrt {x + 2} } \right)$.
For large $x$, we have $\sqrt {x + 1} \approx \sqrt{x}$ and $\sqrt {x + 2} \approx \sqrt{x}$. Thus, $f\left( x \right) \to 0$ as $x \to \infty $. So, $f\left( x \right)$ converges. Therefore, the sequence also converges.
Now, we find the limit of the sequence:
$\mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {n + 1} - \sqrt {n + 2} } \right)$
$ = \mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {n + 1} - \sqrt {n + 2} } \right)\dfrac{{\left( {\sqrt {n + 1} + \sqrt {n + 2} } \right)}}{{\left( {\sqrt {n + 1} + \sqrt {n + 2} } \right)}}$
$ = \mathop {\lim }\limits_{n \to \infty } \dfrac{{n + 1 - \left( {n + 2} \right)}}{{\sqrt {n + 1} + \sqrt {n + 2} }} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{ - 1}}{{\sqrt {n + 1} + \sqrt {n + 2} }} = 0$
Thus, the sequence converges and its limit is $0$.
