## Infinitely Nested Radicals (Square Roots)

Problem: Evaluate $$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$$ . Now evaluate $$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}$$  and  $$\sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}}$$  for  $$n>0$$.

Solution:

Let $$x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$$ . Squaring both sides gives:

$x^2=1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$

$x^2-1=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$

But notice that the RHS is just $$x$$, so:

$x^2-1=x$

$x^2-x-1=0$

To which the solutions are $$x=\frac{1 \pm \sqrt5}{2}$$. The negative solution $$\left(x=\frac{1 - \sqrt{5}}{2}\right)$$ is clearly extraneous, since the radical ($$\sqrt{}$$) denotes the positive square root. Therefore $$x=\frac{1 + \sqrt{5}}{2}$$, which is the golden ratio, $$\phi$$.

We can now get the general solution in a similar fashion. Let $$x=\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}$$ . Squaring both sides gives:

$x^2=n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}$

$x^2-n=\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}$

$x^2-n=x$

$x^2-x-n=0$

$x=\frac{1 \pm \sqrt{1+4n}}{2}$

Ignoring the negative solution, we get that:

$x=\frac{1}{2} (1+\sqrt{1+4n})$

Similarly, let $$x=\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}$$ :

$x^2=n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}$

$x^2-n=-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}$

$x^2-n=-x$

$x^2+x-n=0$

$x=\frac{-1 \pm \sqrt{1+4n}}{2}$

Ignoring the negative solution, we get that:

$x=\frac{1}{2} (-1+\sqrt{1+4n})$