The n-Category Café

Here is Euler’s continued fraction formula:

$$a_0 + a_0a_1 + a_0a_1a_2 + \cdots + a_0a_1a_2\cdots a_n =$$ $$\frac{a_0}{1 - \frac{a_1}{1 + a_1 - \frac{a_2}{1 + a_2 - \frac{\ddots}{\ddots \frac{a_{n-1}}{1 + a_{n-1} - \frac{a_n}{1 + a_n}}}}}}\,$$

The easiest way to convince yourself of it is to take the right-hand side for some smallish value of $n$ and keep simplifying it in obvious ways — putting sums of fractions on common denominators, etc. — until you get the left-hand side.

It’s a nice way to turn finite sums into fractions… and then, by taking a limit, infinite series into continued fractions.

For example, starting with

$$\begin{array}{ccl} \arctan x &=& \displaystyle{ \int_0^x \frac{d t}{1 + t^2} } \\ \\ & = & \displaystyle{ \int_0^x \left(1 - t^2 + t^4 - t^6 \cdots \right) d t } \\ \\ &=& \displaystyle{ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots } \\ \\ &=& x + x \left(\frac{-x^2}{3}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right)\left(\frac{-5x^2}{7}\right) + \cdots \end{array}$$

and then applying Euler’s continued fraction formula, we get

$$\begin{array}{ccl} \arctan x &=& x + x \left(\frac{-x^2}{3}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right) + x \left(\frac{-x^2}{3}\right)\left(\frac{-3x^2}{5}\right)\left(\frac{-5x^2}{7}\right) + \cdots \\ \\ &=& \frac{x}{1 - \frac{\frac{-x^2}{3}}{1 + \frac{-x^2}{3} - \frac{\frac{-3x^2}{5}}{1 + \frac{-3x^2}{5} - \frac{\frac{-5x^2}{7}}{1 + \frac{-5x^2}{7} - \ddots}}}} \\ \\ &=& \frac{x}{1 + \frac{x^2}{3 - x^2 + \frac{(3x)^2}{5 - 3x^2 + \frac{(5x)^2}{7 - 5x^2 + \ddots}}}} \end{array}$$

where in the last step we did a lot of multiplying numerators and denominators by the same odd number.

All this works within the radius of convergence of the geometric series, namely for $|x| < 1$. But with some more care it works at $x = 1$, which is nice because $\arctan 1 = \frac{\pi}{4}$. So we get

$$\frac{\pi}{4} = \frac{1}{1 + \frac{1^2}{3 - 1 + \frac{3^2}{5 - 3 + \frac{5^2}{7 - 5 + \cdots}}}}$$

or in other words

$$\frac{4}{\pi} = 1 + \frac{1^2}{2 + \frac{3^2}{2 + \frac{5^2}{2 + \frac{7^2}{2 + \ddots}}}}$$

You can see other interesting examples of this method on Wikipedia.