4. Given that $\left\{a_{n}\right\}$ is a geometric sequence, $a_{2}=2, a_{5}=\frac{1}{4}$, then the range of $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}\left(n \in \mathbf{N}^{*}\right)$ is . $\qquad$
4. $\left[8, \frac{32}{3}\right)$ Detailed explanation: Let the common ratio of $\left\{a_{n}\right\}$ be $q$, then $q^{3}=\frac{a_{5}}{a_{2}}=\frac{\frac{1}{4}}{2}=\frac{1}{8}$, hence $q=\frac{1}{2}$. Therefore, the sequence $\left\{a_{n} a_{n+1}\right\}$ is a geometric sequence with the first term $a_{1} a_{2}=8$ and the common ratio $q^{2}=\frac{1}{4}$. $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}=\frac{8\left(1-\frac{1}{4^{n}}\right)}{1-\frac{1}{4}}=\frac{32}{3}(1-$ $4^{-n}$ ). Clearly, $8=a_{1} a_{2} \leqslant a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}<\frac{32}{3}$.