What is the value of x = sqrt( 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ...))))?

First we must define the problem rigorously.

Let
                      a(1) = sqrt(2)
                      a(2) = sqrt( 2 + sqrt(2))
                      a(3) = sqrt( 2 + sqrt(2 + sqrt(2)))
                      a(n) = sqrt( 2 + sqrt(2 + ... +sqrt(2)))...)
                                             (n sqrt's)

Let
                        x = lim    a(n)
                            n->inf

Show that the limit (x) exists, and find its value.

We have
                      a(1) = sqrt(2)              (i)
                      a(n+1) = sqrt( 2 + a(n))    (ii)

Therefore             if  a(n) < 2   then
                      a(n+1) < sqrt (2 +2) = 2
but                       a(1) < 2

Therefore by induction    a(n) < 2 for all n      (iii)

by (iii)              a(n)**2 = a(n) * a(n)
                              <  2 * a(n)
                              <  a(n) + a(n)
                              <  2 + a(n)

Therefore by (ii)       a(n+1) > a(n) for all n   (iv)

The a(n)'s form an increasing sequence bounded above (by 2),
so the sequence must have a limit. (i.e. x exists)
Furthermore x must be greater than a(1) = sqrt(2) so:  sqrt(2) < x <= 2.

Now                     a(n+1) = sqrt(2 + a(n)) [Editor: was
                                 "sqrt(2 + sqrt(a(n))"]
                        a(n+1)**2 = 2 + a(n)

Taking the limit as n goes to infinity of each side (this step is now
justified as we know the limit exists) we obtain

                   lim  a(n+1)**2 = 2 + lim  a(n)
                  n->inf               n->inf

                             2
                            x  = 2 + x         (v)

The solutions of (v) are x = 2 and x = -1. But as x > sqrt(2) we must
have x=2.

Note that although x must be a solution of equation (v) it is not true
that any solution to (v) is equal to x.  -1 is called an extraneous root,
that is an extra solution brought about by squaring an equation.
Consider a simple example the equation a=b.  If we square both sides
we obtain a**2=b**2 with solutions a=b and a = -b .  When solving an
equation by squaring both sides one must always be on the lookout for
extraneous roots.  (as indicated in the original posting if we take
both sides of the equation to the fourth power we obtain even more
extraneous roots).

Therefore METHOD 1 is valid but needs more details to be rigorous

METHOD 2 is equivalent to noting that

                     a(n) = 2 cos [ pi/(2**(n+1)) ]   (vi)

Therefore         lim  a(n) = lim  2 cos [ pi/(2**(n+1)) ]
                 n->inf      n->inf

                            = 2 cos(0) = 2

This method is perfectly valid (although to make it rigorous one would
have to prove the identity (vi) ). It is not however as general as method 1
the basic ideas of which can be applied to a very large class of
similar problems.

-- William Hughes (csc@watmath.UUCP)

[ Editor: the original posting, with its Methods 1 and 2, was not
collected. ]