The first in a series is the radical

x = \sqrt{2 + \sqrt{2 + \sqrt{2 + ...}}}

where the expression, i.e., the addition of radicals embedding even more radicals, is understood to continue for ever. The meaning of this expression may be made quite precise as a limit of an infinite sequence of numbers:

x_1 = \sqrt{2}, \quad x_2 = \sqrt{2 + \sqrt{2}}, \quad x_3 = \sqrt{2 + \sqrt{2 + \sqrt{2}}}, ...

The sequence is monotone increasing. Indeed in passing from x_k  to x_{k+1}  we replace a 2 with 2 + \sqrt{2} , making x_k \lt x_{k+1} .

Naturally enough, the Monotone Convergence Theorem assures the existence of the limit x = \lim_{n \to \infty}x_n . To use the theorem we need to show that the sequence x_n  is bounded. This is done by induction. We claim that \forall n\,x_n \le 2 .

Indeed, this is true for x_1 = \sqrt{2} . Assume also that, for some k , x_k \le 2 . Then x_{k+1} = \sqrt{2 + x_k}  so that

x_{k+1}^2 = 2 + x_k \le 4

implying x_{k+1} \le 2  and we are finished with induction.

Now that we know that the sequence x_k  has a limit x  we are also able to find its value. Indeed

x = \sqrt{2 + x}

squaring which we get a quadratic equation x^2 - x - 2 = 0 , with two solutions x = 2  and x = -1 . The latter, being negative, is clearly unsuitable and we are left with the identity: x = 2.

The result is quite pretty

\sqrt{2 + \sqrt{2 + \sqrt{2 + ...}}} = 2.

The next step is to consider the more general expression

x = \sqrt{a + \sqrt{a + \sqrt{a + ...}}},

for some a \gt 0 . We'll have to separate two cases: a \gt 2  and a \lt 2 .


Calculus