5. Given the increasing sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=\frac{2}{3}, a_{n+1}^{2}=a_{n+1}+a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

The sequence $b_{n}=\frac{(-1)^{n-1}}{a_{n}-1}$, the sum of the first $n$ terms is $S_{n}$. Then $\left[S_{2021}\right]$ is $\qquad$
5. -4 .
$$
\begin{array}{l}
\text { Given } a_{n+1}^{2}-a_{n+1}=a_{n} \\
\Rightarrow \frac{1}{a_{n}}=\frac{1}{a_{n+1}^{2}-a_{n+1}}=\frac{1}{a_{n+1}-1}-\frac{1}{a_{n+1}} \\
\Rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n}}+\frac{1}{a_{n+1}} .
\end{array}
$$

Given $b_{n}=\frac{(-1)^{n-1}}{a_{n}-1}$, then
$$
\begin{array}{l}
S_{2021}=\frac{1}{a_{1}-1}-\frac{1}{a_{2}-1}+\frac{1}{a_{3}-1}-\cdots+\frac{1}{a_{2021}-1} \\
=\frac{1}{a_{1}-1}-\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}\right)+\left(\frac{1}{a_{2}}+\frac{1}{a_{3}}\right)-\cdots- \\
\left(\frac{1}{a_{2019}}+\frac{1}{a_{2020}}\right)+\left(\frac{1}{a_{2020}}+\frac{1}{a_{2021}}\right) \\
=\frac{1}{a_{1}-1}-\frac{1}{a_{1}}+\frac{1}{a_{2021}} \\
=-\frac{9}{2}+\frac{1}{a_{2021}} .
\end{array}
$$

Given $a_{1}=\frac{2}{3}$, and $\left\{a_{n}\right\}$ is an increasing sequence, then
$$
a_{2021} \in\left(\frac{2}{3}, 2\right) \Rightarrow \frac{1}{a_{2019}} \in\left(\frac{1}{2}, \frac{3}{2}\right) \text {. }
$$

Therefore, $-4<-\frac{9}{2}+\frac{1}{a_{2021}}<-3$.
Thus, $\left[S_{2021}\right]=-4$.