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Define a sequence $\langle a(n)\rangle$ recursively by $a(1)=\sqrt{2}$ and $a(n+1)=\sqrt{2+a(n)}$ $(n>0)$. a)by induction or otherwise show that the sequence is increasing and bounded above 3. Apply the monotonic sequence theorem to show that the limit as $n$ approaches infinity exists. b)What is the value of the limit as n approaches infinity of the sequence?

I know that to show that a sequence is bounded above I have to show that there exists a real number $M$ (in this case 3) such that $a(n)\leq M$ for all $n\geq1$. The theorem states that if it is bounded above and non-decreasing then it is convergent, since it is increasing then by the theorem it will converge. Does this mean it will converge to 3? But I am thinking it could converge to a number less than three possibly, but could not converge to anything greater than 3.

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    $\begingroup$ Your thinking is right. The limit must be $\le 3$, but need not be $3$. And in fact the limit is $\lt 3$. $\endgroup$ – André Nicolas May 17 '15 at 6:32
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Once you have proved that the sequence is convergent, then, making $n \to \infty$ in the recurrence relation defining $a_n$, the limit $\ell$ satisfies $$ \ell =\sqrt{2+\ell}, \quad \ell>0. $$ equivalently $$ \ell^2 =2+\ell, \quad \ell>0. $$ which may be solved to obtain $\ell$.

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suppose lim$a_n$=x, Then lim$a_n$+$_1$=x also

now $x=√(2+x)$ squaring it you will get $x^2$=$x+2$

solving this quadratic equation we get $x$=-1,2

$x$=-1 is absurd as sequence is increasing and starts from $√2$

we have only choice $x=2$, which is the limit of the sequence

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