how to prove {Sn} converges, if S1=sqrt(2), S(sub n+1) = sqrt(2+sqrt(Sn))?

Answer 1

S1 = sqrt(2), so s1 < 2.

S2 = sqrt(2+S1) < sqrt(2+2) = sqrt(4) = 2

S3 = sqrt(2+S2) < sqrt(4) = 2

etc. For every Sn, it is easy enough to show that Sn < 2.

On the other hand, S1 = sqrt(2) > 0.

S2 = sqrt(2+S1) > sqrt(2) = S1.

S3 = sqrt(2+S2) > sqrt(2+S1) = S2.

Therefore, you may show that Sn > S(n-1).

When you combine those two facts: Sn > S(n-1) and Sn < 2 for all n, then you have shown that the sequence converges; in fact, it converges to the value 2.

Answer 2

The previous answer is wrong. That person worked with the recurrence relation
S(n+1) = sqrt(2 + S(n))

You do need to establish that the sequence is bounded.

What you want to do next is show that the sequence is monotone. You can do this if you can establish that S(n+1) - S(n) is always positive, or else always negative. Do this by induction.

For the general step in the induction, replace both S(n+1) and S(n) by their equivalents from the recurrence relation, then rewrite using the fact that
sqrt(a) - sqrt(b) = (a - b)/(sqrt(a) + sqrt(b))