China Girls' Mathematical Olympiad

Let $a_0=1$. We prove the following identity: $$\sum _{k=1}^n\prod _{j=1}^k\frac{a_{j-1}}{1+a_j}=1-\prod _{j=1}^n\frac{a_j}{1+a_j}\stackrel{◯}{1}$$ by induction on $n$.

It is evident that $\stackrel{◯}{1}$ is true for $n=1$. Suppose that $\stackrel{◯}{1}$ is true for $n-1$, $n\ge 2$, then for $n$, $$\begin{array}{rl}\sum _{k=1}^n\prod _{j=1}^k\frac{a_{j-1}}{1+a_j}& =\sum _{k=1}^{n-1}\prod _{j=1}^k\frac{a_{j-1}}{1+a_j}+\prod _{j=1}^n\frac{a_{j-1}}{1+a_j}\\ & =1-\prod _{j=1}^{n-1}\frac{a_j}{1+a_j}+\prod _{j=1}^n\frac{a_{j-1}}{1+a_j}\\ & =1-\prod _{j=1}^n\frac{a_j}{1+a_j}.\end{array}\square$$