# Nested fractional denominator(up to infinity)calculation Question: 4 + (1)/(2 + (1)/(1 + (1)/(3 + (1)/(1+(1)/(2 + (1)/(8 + (1)(...))))))) = sqrt(A) Find the positive integer A in the equation above.

Nested fractional denominator(up to infinity)calculation
Question:
$4+\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{8+\frac{1}{\ddots }}}}}}}=\sqrt{A}$
Find the positive integer A in the equation above.
Details and assumptions
The pattern repeats 2, 1, 3, 1, 2, 8 infinitely, but 4 comes only once, i.e., at the beginning.
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Willie Smith
Let's take a simpler example first: say,
$\sqrt{A}=1+\frac{2}{3+\frac{2}{3+...}}.$
Here the repetition $2,3,2,3,...$ continues forever after the $1$.
The problem is that this looks on the face of it like some sort of "infinitary" equation (or rather, it's a normal equation with an "infinitary" term in it). We want to somehow "tame" the right-hand side.
This is where the repetition, or self-similarity, comes in. Let $x=\frac{2}{3+\frac{2}{3+...}}$. Then if we take the reciprocal of both sides, we see something interesting:
$\frac{1}{x}=\frac{3+x}{2}.$
Solving for x yields
${x}^{2}+3x-2=0,$
which has as its solutions
$\frac{-3±\sqrt{9+8}}{2,}$
that is,
$x=\frac{\sqrt{17}-3}{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x=\frac{-\sqrt{17}-3}{2}.$
Since x is clearly positive∗, it has to be the former.
Plugging back into the original expression, we have
$\sqrt{A}=1+x=\frac{\sqrt{17}-1}{2}.$
Squaring both sides finishes the problem.
Can you see how to follow a similar argument here? (It's definitely going to be messier since your patter takes longer to repeat, but the basic idea still applies.)
An interesting feature of this argument is: it totally breaks down if we don't have repetition! To see what I mean, try to calculate
$\frac{1}{2+\frac{3}{4+\frac{5}{6+...}}.}$
In general, things are much more complicated and the "naive" argument above simply doesn't work.