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:

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

(0 Votes)