Solution

The finite simple continued fractions $[1;2]$, $[1;2,2]$, $[1;2,2,2]$, $[1;2,2,2,2]$, $[1;2,2,2,2,2], \ldots $ are called the convergents of $[1;\overline{2}]$. More generally, the $n^{th}$ convergent for a given continued fraction is found by truncating the original continued fraction sequence to $n$ values.

First, find the exact value of $[1;\overline{2}]$. Then, to get a feel for how closely the convergents approximate their limiting value, find the difference between $[1;\overline{2}]$ and the value of each convergent listed.


First, we note that $[1;\overline{2}]$ has a fractional part of

$$y = \frac{1}{2 + \displaystyle{\frac{1}{2 + \displaystyle{\frac{1}{2 + \cdots}}}}}$$

This tells us that

$$y = \frac{1}{2+y}$$

Multiplying by $(2+y)$, we produce a quadratic equation

$$2y+y^2 = 1$$

Putting this in standard form and solving for $y$ with the quadratic formula, we find

$$y^2 + 2y - 1 = 0 \quad \quad \longrightarrow \quad \quad y = -1 \pm \sqrt{2}$$

Note, we can throw away one of the solutions found, as we know $y$ must be positive. Hence,

$$y = -1 + \sqrt{2}$$

This in turn, tells us that $[1;\overline{2}] = 1 + y = 1 + (-1 + \sqrt{2}) = \sqrt{2}$.

Now we turn our attention to the convergents...

Through direct calculation, we find

$$\begin{array}{rcccl} [1;2] &=& 3/2 &=& 1.5\\ [1;2,2] &=& 7/5 &=& 1.4\\ [1;2,2,2] &=& 17/12 &=& 1.41\overline{6}\\ [1;2,2,2,2] &=& 41/29 &=& 1.41379310\cdots\\ [1;2,2,2,2,2] &=& 99/70 &=& 1.41428571\cdots\\ \end{array}$$

We compare these with $\sqrt{2} = 1.41421356\cdots$, and see that the sequence of convergents converges fairly quickly.

Interestingly, rather than evaluate each successive convergent from scratch, we can tighten the calculation process by using each convergent value to find the next convergent value.

Specifically, if $a_n$ is the $n^{th}$ convergent of $[1,\overline{2}]$, then

$$a_{n+1} = \frac{1}{a_n + 1} + 1$$

Do you see where this comes from? Write down the first couple of convergents in full continued fraction form, and look carefully at the denominators. How can you modify one convergent's form to produce the next one?