How to solve "nested" problems

Something that I see a lot at math competitions or on Brilliant is a problem asking to find the value of $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ for $x=$ some value. I'm going to talk about how to solve that "nested radical" as well as "nested" fractions.

First I'm going to talk about how to solve a problem such as finding the value of $\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}.$ Let $k=\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}.$ You may notice that you are adding $k$ to $6$ and taking the square root of this. I'll demonstrate this with some parentheses. $\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\sqrt{6+\left(\sqrt{6+\sqrt{6+\ldots}}\right)}=\sqrt{6+\sqrt{\left(6+\sqrt{6+\ldots}\right)}}=\ldots$ For simplicity, we're just going to use $\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\sqrt{6+\left(\sqrt{6+\sqrt{6+\ldots}}\right)}.$ You may notice that this is the same thing as saying that $k=\sqrt{6+k}.$ Let's try solving that. $\begin{aligned} k&=\sqrt{6+k}\\ k^2&=6+k\\ k^2-k-6&=0\\ (k-3)(k+2)&=0\\ k=3\text{ or }k=-2\\ \end{aligned}$ We can eliminate the $k=-2$ solution because the nested radical obviously is non-negative. Thus, we can say that $\sqrt{6+\sqrt{6+\sqrt{6+\ldots}}}=\boxed{3}.$

Now we can use a similar line of logic to find a general form for $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}.$ Let that equal $y.$ We will make use of the same procedure, but using the quadratic formula to solve for $y.$ $\begin{aligned} y&=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}\\ y&=\sqrt{x+y}\\ y^2&=x+y\\ y^2-y-x&=0\\ y&=\dfrac{1\pm\sqrt{1+4x}}{2}\\ \end{aligned}$ Since we are dealing with real numbers, the value under the radical cannot be negative (except for $x=0$.) Thus, we can say that $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}=\dfrac{1+\sqrt{1+4x}}{2}.$

Now let's use the same logic to derive a formula for nested fractions of the form $\dfrac{1}{a+\frac{1}{a+\ddots}}$ as well as continued fractions where the numerator is not necessarily $1.$

First, the obligatory demonstration of such a fraction. $\begin{aligned} \dfrac{1}{2+\frac{1}{2+\ddots}}&=\dfrac{1}{2+\left(\frac{1}{2+\ddots}\right)}\\ k&=\dfrac{1}{2+k}\\ k^2+2k&=1\\ k^2+2k-1&=0\\ k&=\dfrac{-2\pm\sqrt{4+4}}{2}\\ k&=-1\pm\sqrt{2} \end{aligned}$ Again, we can throw out the negative solution and say that $\dfrac{1}{2+\frac{1}{2+\ddots}}=-1+\sqrt{2}.$

Moving on, we can derive a general formula for finding $\dfrac{c}{x+\frac{c}{x+\ddots}},$ allowing this expression to be equal to $y.$

$\begin{aligned} y&=\dfrac{c}{x+\frac{c}{x+\ddots}}\\ y&=\dfrac{c}{x+y}\\ y^2+xy&=c\\ y^2+xy-c&=0\\ y&=\dfrac{-x+\sqrt{x^2+4c}}{2} \end{aligned}$

So a summary of what we have proved here:

• $\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}=\dfrac{1+\sqrt{1+4x}}{2}$
• $\dfrac{c}{x+\frac{c}{x+\ddots}}=\dfrac{-x+\sqrt{x^2+4c}}{2}$

If you want some practice problems, here's two.

1. Find a closed form for $x+\dfrac{1}{2x+\frac{1}{2x+\ddots}}.$

2. This calculus problem

3. Find all values of $n$ for which $\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$ is an integer.

I hope this helps you out! Please feel free to share what you think in the comments. Thank you very much!