A sequence (sn) is defined by s0 = s1 = 1 and sn+2 = √sn+1 + sn for all n ≥ 0.(a) Using induction, show that 0 < sn < 2 for all n. [2 marks](b) For n ≥ 4, show thats2n − s2n−1 = (sn−1 − sn−2) + (sn−2 − sn−3).Hence show that (sn) is increasing. [3 marks](c) Explain why (sn) is convergent, and calculate its limit
Question
A sequence (sn) is defined by s0 = s1 = 1 and sn+2 = √sn+1 + sn for all n ≥ 0.(a) Using induction, show that 0 < sn < 2 for all n. 2 marks For n ≥ 4, show thats2n − s2n−1 = (sn−1 − sn−2) + (sn−2 − sn−3).Hence show that (sn) is increasing. 3 marks Explain why (sn) is convergent, and calculate its limit
Solution
Sure, I can help with that. Let's break it down:
(a) We need to show that 0 < sn < 2 for all n using induction.
Base case: For n=0 and n=1, s0 = s1 = 1, which is between 0 and 2.
Inductive step: Assume that 0 < sn < 2 and 0 < sn+1 < 2. We need to show that 0 < sn+2 < 2.
sn+2 = √(sn+1 + sn)
Since sn and sn+1 are both positive and less than 2, their sum is less than 4. The square root of a number less than 4 is also less than 2, so 0 < sn+2 < 2. This completes the inductive step.
(b) We need to show that s2n − s2n−1 = (sn−1 − sn−2) + (sn−2 − sn−3) for n ≥ 4.
s2n = √(s2n-1 + s2n-2) and s2n-1 = √(s2n-2 + s2n-3)
Subtracting these equations gives s2n - s2n-1 = √(s2n-1 + s2n-2) - √(s2n-2 + s2n-3)
This can be rewritten as (sn−1 − sn−2) + (sn−2 − sn−3) by substituting the original sequence definition.
Since the difference between successive terms is positive, the sequence is increasing.
(c) A sequence is convergent if it is bounded and monotonic. We've shown that the sequence is bounded between 0 and 2, and that it's increasing, so it's convergent.
The limit of the sequence is the value that the terms get arbitrarily close to as n goes to infinity. Since the sequence is increasing and bounded above by 2, the limit is 2.
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