Euler's continued fraction formula

Remarks

Proof idea: by induction on $n$. Replacing $a_n$ by $a_n(1 + a_{n+1})$ turns the left side for $n$ into the left side for $n + 1$. On the right side the same replacement turns the last fraction $\frac{a_n}{1 + a_n}$ into $\frac{a_n}{1 + a_n - \frac{a_{n+1}}{1 + a_{n+1}}}$, because both equal $\frac{a_n(1 + a_{n+1})}{1 + a_n + a_n a_{n+1}}$. Check for $n = 1$: $\cfrac{a_0}{1 - \frac{a_1}{1 + a_1}} = a_0(1 + a_1)$. Example: with $a_0 = 1$ and $a_k = x$ for $k \ge 1$, the series is the geometric series, and for $\abs{x} < 1$ $$\frac{1}{1 - x} = \cfrac{1}{1 - \cfrac{x}{1 + x - \cfrac{x}{1 + x - \cdots}}}.$$

Going deeper

For a student

Power series have this form with $a_k = c_k x / c_{k-1}$ when the coefficients $c_k$ are nonzero. In this way the formula turns the series of $e^x$, $\log(1 + x)$ or $\arctan x$ into continued fractions. These continued fractions converge at the same rate as the series; other transformations are needed to obtain faster convergence.