If $\alpha \in (0,1)$ is an irrational number, the question is to show that
\begin{equation} \alpha = \sum_{n=0}^{\infty} \frac{(-1)^n}{q_nq_{n-1}} \end{equation}
(where $q_n$ denotes the denominator of the n$^{th}$ partial quotient of $\alpha$).
What I have tried to use is that if:
\begin{equation} p_nq_{n-1} - q_np_{n-1} = (-1)^{n+1}\\ \frac{p_nq_{n-1}}{q_n} - p_{n-1} = \frac{(-1)^{n+1}}{q_n}\\ \frac{p_n}{q_n} - \frac{p_{n-1}}{q_{n-1}} = \frac{(-1)^{n+1}}{q_nq_{n-1}}\\ \end{equation}
Since, by definition, $\alpha = \lim_{n\rightarrow \infty}\frac{p_n}{q_n}$, it looks like it could be a telescopic series using what I have calculated. Does anyone know if I'm on the right track here?
Update: For example, if I re-write the equation above;
\begin{equation} \frac{p_n}{q_n} = \frac{(-1)^{n+1}}{q_nq_{n-1}} + \frac{p_{n-1}}{p_{n-1}}\\ \frac{p_n}{q_n} = \frac{(-1)^{n+1}}{q_nq_{n-1}} + \frac{(-1)^n}{q_{n-1}q_{n-2}} + \frac{p_{n-2}}{q_{n-3}}\\ \end{equation}