Show that √7, √(7 - √7), √(7 - √(7 + √7)), √(7 - √(7 + √(7 - √7))), ... converges and find its limit.

**Solution**

Answer: 2.

Let x_{n} be the nth number in the sequence. We have x_{n+2} = √(7 - √(7 + x_{n})). Hence if x_{n} < 2, then x_{n+2} > 2, and if x_{n} > 2, then x_{n+2} < 2. So if the sequence converges, its limit must be 2.

If x_{n} = 2 + ε, with 0 < ε < 1, then 9 + ε < 9 + ε + ε^{2}/36 = (3 + ε/6)^{2}, so x_{n+2} > √(7 - (3 + ε/6)) = √(4 - ε/6). But certainly ε/6 < ε/3 - ε^{2}/144, so √(4 - ε/6) > 2 - ε/12. Thus x_{n+2} differs from 2 by less than ε/12. Similarly, if x_{n} = 2 - ε, with 0 < ε < 1, then 9 - ε > 9 - ε/2 + ε^{2}/144 = (3 - ε/12)^{2}, so x_{n+2} < √(4 + ε/12) < 2 + ε/48.

So we have established that if |x_{n} - 2| < 1, then |x_{n+2} - 2| < 1/12 |x_{n} - 2|. But we certainly have |x_{1} - 2| < 1, and |x_{2} - 2| < 1, so x_{n} converges to 2.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002