#### Answer

a) The sequence is increasing and has an upper bound.
b) $\lim\limits_{n \to \infty}a_{n} = 2$

#### Work Step by Step

a)
$a_{1}= \sqrt 2$
$a_{n+1} = \sqrt (2+a_{n})$
For $n=1$
$a_{1+1} = \sqrt (2+a_{1})$
$a_{2}= \sqrt (2+\sqrt 2)$ since $\sqrt (2+\sqrt 2) \gt \sqrt 2$
Then $a_{2} \gt a_{1}$
which suggests that the sequence is increasing
Assume that it is true for $n=k$
so $a_{k+1} \gt a_{k}$
$2 + a_{k+1} \gt 2+a_{k}$
$\sqrt (2+a_{k+1}) \gt \sqrt (2+a_{k})$
$a_{k+2} \gt a_{k+1}$
The sequence is increasing for $n=k+1$
$a_{n+1} \gt a_{n}$ for all $n$
For the sequence {$a_{n}$} is increasing.
$a_{1} = \sqrt 2 = 1.4142$
$a_{2} = \sqrt (2+\sqrt 2) = 1.84776$
$a_{3} = \sqrt (2+a_{2}) =1.96157$
$a_{4} = 1.990$
$a_{5} = 1.9976$
$a_{6}=1.999$
$a_{10}=1.9999$
The terms are approaching to $2$
$2 \lt 3$
So $a_{n} \lt 3$ for all $n \geq 1$
Thus 3 is an upper bound.
b)
Since {$a_{n}$} is increasing and bounded above by 4 then {$a_{n}$} must have a limit.
Let limit be $L$
$\lim\limits_{n \to \infty}{a_{n}} = L$ exists
$\lim\limits_{n \to \infty} a_{n+1} = \lim\limits_{n \to \infty} \sqrt (2+a_{n})$
$= \sqrt (2+\lim\limits_{n \to \infty}a_{n})$
$=\sqrt (2+L)$
Since $a_{n} → L$ then $a_{n+1} → L$ (as $n→ \infty$)
$L = \sqrt (2+L)$
$L^{2} = 2+L$
$L^{2}-L-2=0$
$(L-2)(L+1)=0$
$L=-1$ or $L=2$
since
$a_{1} = \sqrt 2 \gt 0$ and $a_{n}$ is increasing so $L \ne -1$
and thus
$\lim\limits_{n \to \infty}a_{n} = 2$