Asequence{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.

It is clear that a(n)=2^(1-2^(-n)). In fact, for n=1 this produces 2^(1-1/2)=sqrt(2)=a1 and if it is true for a(n) then a(n+1) = sqrt (2 * 2^(1-2^(-n))) = sqrt(2^(2-2^(-n))) = 2^(1-2^(-(n+1))) (a) clearly 2^(1-2^(-n))<2<3 so the sequence is bounded by 3. Also a(n+1)/a(n) = 2^(1-2^(-n-1) - 1+2^(-n)) = 2^(1/2^n - 1/2^(n+1)) = 2^(1/2^(n+1)) >1 so the sequence is monotonically increasing. As it is monotonically increasing and has an upper bound it means it has a limin when n-> oo (b) 1-1/2^n -> 1 as n->oo so 2^(1-2^(-n)) -> 2 as n->oo

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