Finding the limit of a sequence.

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SUMMARY

The sequence \{s_n\} defined by the recurrence relation s_{n+1} = \frac{1}{2} (s_n + s_{n-1}) is proven to be convergent, with the limit being the average of the initial terms s_1 and s_2. The sequence is bounded between s_1 and s_2, but it is neither monotonically increasing nor decreasing. The Cauchy convergence test is applicable here, demonstrating that the distance between consecutive terms shrinks by half with each iteration, confirming convergence.

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Homework Statement



A sequence \{s_n\} is defined by s_{n+1} = \frac{1}{2} (s_n + s_{n-1}); s_1 > s_2 > 0 I have to prove that the sequence is convergent and I have to find the limit.

Homework Equations





The Attempt at a Solution


I tried equating the limit of both sides to get s = (1/2)(s + s) but then I just get s = s. I managed to find that the sequence is bounded between s_1 and s_2 but the sequence is neither increasing nor decreasing.

For proving the convergence I tried used the Cauchy convergence test.
|\frac{1}{2}(s_n + s_{n-1}) - s_n| = |\frac{1}{2}(s_{n-1} - s_n)| < \in is as far as I got. I can't assume that the s_n tends to a limit s because I haven't proved it does yet. So I'm stuck.

I'm also completely stuck on how to prove the convergence of a sequence defined by a recurrence relation in which the relation involves TWO of the previous terms.

where s_n = f(s_n, s_{n-1}).
 
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(s_n+ s_{n-1})/2 is exactly half way between s_n and s_{n-1} That means that the distance between two consecutive terms is half the distance between the preceding termS: |s_n- s_{n-1}|= (1/2)|s_{n-1}- s_{n-2}|
 

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