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

Keegan

English

8 September, 04:13

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.

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Answers (1)

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Ben Daniel · Answer 1

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)) <21 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

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.

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.