A sequence{an} is given by a1=sqrt(2), an+1=sqrt(2+an).
a) by induction or otherwise, show that {an} is increasing and bounded above by 3. Apply the Monotonic Sequence Theorem to show that lim n--infinity an exists.
b) Find lim n--infinity an.

Question

A sequence{an} is given by a1=sqrt(2), an+1=sqrt(2+an).
a) by induction or otherwise, show that {an} is increasing and bounded above by 3. Apply the Monotonic Sequence Theorem to show that lim n-->infinity an exists.
b) Find lim n-->infinity an.

anonymous · Answer

$$a_{n+1}=\sqrt{a_n+2}$$

anonymous · Answer

a) in order for $a_n$ to be increasing we must show that $a_{n+1}>a_n$ for  $n\ge1$

anonymous · Answer

u r offline so i go to give a complete solution

mathslover · Answer

No need to give complete soln @mukushla

anonymous · Answer

first show that  $a_n<2$ by mathematical induction :

Proof: For each positive integer $n\ge1$ , let  $S(n)$  be the statement  $a_n<2$

Basis step: $S(1)$ is the statement $\sqrt{2}<2$ . Thus $S(1)$ is true. 

Inductive step: We suppose that $S(k)$ is true and prove that $S(k+1)$ is true. 

Thus, we assume that
$a_k<2$

and prove that
$a_{k+1}<2$

we have

$a_{k+1}=\sqrt{a_k+2}<\sqrt{2+2}=2$

hence proved

mathslover · Answer

You did it :(

anonymous · Answer

o sorry

anonymous · Answer

i think thats against the code of contact...ok for other parts i will just give some hints

anonymous · Answer

in order to showing the sequence is increasing :$$a_{n+1}-a_n=\sqrt{a_n+2}-a_n=\frac{(\sqrt{a_n+2}-a_n)(\sqrt{a_n+2}+a_n)}{(\sqrt{a_n+2}+a_n)}=\frac{a_n+2-a_n^2}{(\sqrt{a_n+2}+a_n)}$$show that the later expression>0 using the result of induction

anonymous · Answer

i think u got the other parts from here :)