We wanted to figure out the value of the fathomless fraction. We can start by saying the value of this fraction equals x1. x1 = 1 + 1 / x2, and so on. We can define this a recursive sequence, where xn = n + 1 / xn+1. But this doesn’t help if we don’t know any of the later terms. It approaches infinity, of course. xn>n, but xn<n+1. Maybe we could figure out the value of a term based on the previous one? xn-1 = n-1 + 1 / xn. Subtract n-1 from both sides. xn-1 -n +1 = 1 / xn. Take the reciprocal of both sides to find that xn = 1 / xn-1-n+1. We can make an alternate version of the fathomless fraction, call it fathomless equivalent fraction, or FEF for short. We soon realize that both lead to the exact same fraction. But x1 = x-1+1, and so xn=x-n+1, and we can make another fraction. But it returns to x1! It’s like with √x+1, it cannot be simplified. The answer is simply ≈1.433.
x_1=1+\frac1{2+\frac1{3+\frac1{4+…}}}\\[16pt] x_n=n+\frac1{x_{n+1}}\\[16pt] x_n=\frac1{x_{n-1}-n+1}\\[16pt] x_1=\frac1{\frac1{1+x_{-1}}}\\[16pt] x_n=x_{-n}+n\\[16pt] x_1=1+\frac1{2-2+\frac1{x_1-1}}=x_1\\[16pt] x_1 \approx 1.433