Prove: Let $H=\frac{A_{1}(n)+A_{3}(n)+A_{5}(n)+\cdots}{1+A_{2}(n)+A_{4}(n)+\cdots}$, it is easy to prove the identity $H=$
$$\frac{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)-\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)+\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)}$$

Notice that $\left|a_{i}\right|<1, \prod_{i=1}^{n}\left(1-a_{i}\right)$ $>0$, thus we have
$$\begin{array}{l} H=\frac{\prod_{i=1}^{n}\left(1+a_{i}\right)-\prod_{i=1}^{n}\left(1-a_{i}\right)}{\prod_{i=1}^{n}\left(1+a_{i}\right)+\prod_{i=1}^{n}\left(1-a_{i}\right)} \\ \frac{\prod_{i=1}^{n}\left(1+a_{i}\right)-\prod_{i=1}^{n}\left(1-a_{i}\right)-2 \prod_{i=1}^{n}\left(1+a_{i}\right)}{\prod_{i=1}^{n}\left(1+a_{i}\right)+\prod_{i=1}^{n}\left(1-a_{i}\right)}=-1 \end{array}$$

Therefore, $|H|<1$.