(14) The sequence $\left\{a_{n}\right\}$ satisfies $a_{n+1}=\frac{1}{2-a_{n}}$. If for any positive integer $n$, $a_{n+1}>a_{n}$, then the range of values for $a_{1}$ is $\qquad$.
(14) From the problem, we know
$$
a_{n+1}-a_{n}=\frac{1}{2-a_{n}}-a_{n}=\frac{\left(a_{n}-1\right)^{2}}{2-a_{n}}>0 \Leftrightarrow a_{n}<1 \Leftrightarrow a_{n+1}=\frac{1}{2-a_{n}}a_{n}$. If $a_{1}>1$, then there exists a positive integer $m$, such that $a_{1}>\frac{m+1}{m}$. In fact, if $a_{n}>\frac{p+1}{p}$, then
$$
2-a_{n}<\frac{p}{p-1} .
$$

Thus, from $a_{1}>\frac{m+1}{m}$, we can get $a_{m}>\frac{2}{1}=2$, which means $a_{m+1}<a_{m}$. Therefore, the range of $a_{1}$ is $a_{1}<1$.