I am trying to find the limit if it exists for the following unending continued fraction: $$1+{1\over{2+{1\over2+{1\over{2+...}}}}}$$
I have discovered this is the continued fraction for $\sqrt2$, but I am unsure how to show this.
Also I am being asked to assume that the even-order and odd-order truncations have a limit as well and calculate each of them. How would I go about finding any of these limits?
On
Let $x = 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}$. Then \begin{align*} x - 2 &= \frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}} \\ \frac{1}{x-2} &= 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}} \\ \frac{1}{x-2} &= x \end{align*} We find two roots to this quadratic equation, $1+\sqrt{2}$ and $1-\sqrt{2}$. Since $x = 2+\dots > 2$, we ignore the negative root and continue with $x = 1+\sqrt{2}$. The number you have is therefore $$1+\frac{1}{x} = 1 + \frac{1}{1+\sqrt{2}} = 1+\frac{1-\sqrt{2}}{1-2} = 1+\sqrt{2}-1 = \sqrt{2}$$
Neglecting the 1 that's added in the beginning, you can find the value of the continued fraction by solving the equation $x = \frac{1}{2 + x}$.