20. Let the sequence $\left\{a_{n}\right\}$ be defined as
$$
a_{1}=a, a_{n+1}=1+\frac{1}{a_{1}+a_{2}+\cdots+a_{n}-1}(n \geqslant 1)
$$

Find all real numbers $a$ such that $0<a_{n}<1(n \geqslant 2)$.
20. From the given conditions, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n-1}=\frac{a_{n}}{a_{n}-1} . \\
\text { Also, } a_{n+1}=\frac{a_{1}+a_{2}+\cdots+a_{n}}{a_{1}+a_{2}+\cdots+a_{n}-1} \\
=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}=\frac{a_{n}^{2}}{\left(a_{n}-\frac{1}{2}\right)^{2}+\frac{3}{4}}(n \geqslant 2) .
\end{array}
$$

Since \( a_{1}=a, a_{2}=\frac{a_{1}}{a_{1}-1} \), therefore, when \( a_{1}=a \neq 0 \), \( a_{n}>0(n \geqslant 3) \), and when \( 0<a_{2}=\frac{a}{a-1}<1 \), i.e., \( a<0 \), we have
$$
\begin{array}{l}
a_{n+1}-a_{n} \\
=\frac{1}{a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}-1}-\frac{1}{a_{1}+a_{2}+\cdots+a_{n-1}-1} \\
<0(n \geqslant 2),
\end{array}
$$

i.e., \( a_{n+1}<a_{n}(n \geqslant 2) \).
Thus, \( 0<a_{n}<1(n \geqslant 2) \).
Therefore, \( a<0 \).