4. Given a sequence of positive terms $\left\{a_{n}\right\}$ satisfying $a_{n}^{n}+a_{n}-1=0$ $\left(n \in \mathbf{N}^{*}\right)$. Below are 5 statements: (1) $0<a_{n} \leqslant$ $\frac{n}{n+1}$; (2) $\frac{n}{n+1} \leqslant a_{n}<1$; (3) $\left\{a_{n}\right\}$ is an increasing sequence; (4) $\left\{a_{n}\right\}$ is a decreasing sequence; (5) $\left\{a_{n}\right\}$ is not a monotonic sequence. The correct statements are
A. (1)(3)
B. (1)(4)
C. (2)(3)
D. (2)(5)
4. A Solution: Clearly, when $01$, $\frac{a_{n}^{n}+\overbrace{1+1+\cdots+1}^{n-1}}{n}>a_{n}$, i.e., $a_{n}^{n}>n a_{n}-(n-1)$, thus, $a_{n}^{n}+a_{n}-1>(n+1) a_{n}-n, 0<a_{n}<\frac{n}{n+1}$. Also, $a_{1}=$ $\frac{1}{2}$, so, $0<a_{n} \leqslant \frac{n}{n+1}\left(n \in \mathbf{N}^{*}\right)$.

If for some positive integer $n$, $a_{n} \geqslant a_{n+1}$, then $a_{n}^{n \neq 1}<a_{n+1}^{n}$ $\leqslant a_{n}^{n}$, hence
$$
a_{n+1}^{n+1}+a_{n+1}<a_{n}^{n}+a_{n+1} \leqslant a_{n}^{n}+a_{n},
$$

which contradicts $a_{n}^{n+1}+a_{n+1}=1=a_{n}^{n}+a_{n}$. Therefore, for any $n$, $a_{n}<a_{n+1}$.