For any integer $k$, since $0<a_{k}<1$, we set $s_{k}=a_{1}+a_{2}+\ldots+a_{k}$ and $b_{k}=a_{k} s_{k-1}^{2} /\left(1-a_{k}\right)$, with the convention that $s_{0}=0$. We will prove the inequality

$$b_{k}<\left(s_{k}^{3}-s_{k-1}^{3}\right) / 3$$

This follows from the fact that

$$\text { (1) } \begin{aligned} & \Leftrightarrow 0<\left(1-a_{k}\right)\left(\left(s_{k-1}+a_{k}\right)^{3}-s_{k-1}^{3}\right)-3 a_{k} s_{k-1}^{2} \\ & \Leftrightarrow 0<\left(1-a_{k}\right) a_{k}\left(3 s_{k-1}^{2}+3 s_{k-1} a_{k}+a_{k}^{2}\right)-3 a_{k} s_{k-1}^{2} \\ & \Leftrightarrow 0<3 a_{k} s_{k-1}^{2}-3 a_{k}^{2} s_{k-1}^{2}+3\left(1-a_{k}\right) s_{k-1} a_{k}^{2}+\left(1-a_{k}\right) a_{k}^{3}-3 a_{k} s_{k-1}^{2} \\ & \Leftrightarrow 0<3\left(1-a_{k}-s_{k-1}\right) s_{k-1} a_{k}^{2}+\left(1-a_{k}\right) a_{k}^{3} \\ & \Leftrightarrow 0<3\left(1-s_{k}\right) s_{k-1} a_{k}^{2}+\left(1-a_{k}\right) a_{k}^{3}, \end{aligned}$$

the last inequality being indeed correct. We conclude that

$$b_{1}+b_{2}+\ldots+b_{n}<\left(s_{n}^{3}-s_{0}^{3}\right) / 3=1 / 3 .$$