## Calculus 8th Edition

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