4. Let the sequence of positive numbers be $\left\{a_{n}\right\}$, with the sum of the first $n$ terms being $S_{n}$, and there exists a positive number $t$, such that for all natural numbers $n$, $\sqrt{t S_{n}}=\frac{t+a_{n}}{2}$ holds. If $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}<t$, then the range of values for $t$ is
4. $t>\frac{\sqrt[3]{2}}{2}$.

From $\sqrt{t S_{n}}=\frac{t+a_{n}}{2}$, we get $S_{n}=\frac{1}{4 t}\left(t+a_{n}\right)^{2}$.
When $n=1$, $a_{1}=S_{1}=\frac{1}{4 t}\left(t+a_{1}\right)^{2}$, we get $a_{1}=t$;
When $n \geqslant 2$,
$$
a_{n}=S_{n}-S_{n-1}=\frac{1}{4 t}\left[\left(t+a_{n}\right)^{2}-\left(t+a_{n-1}\right)^{2}\right] \text {. }
$$

Simplifying, we get $a_{n}-a_{n-1}=2 t$. Therefore, $a_{n}=(2 n-1) t$.
$$
\therefore S_{n}=n^{2} t \text {. }
$$

From $\lim _{n \rightarrow \infty} \frac{\sqrt{S_{n}}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{n \sqrt{t}}{(2 n-1) t}=\frac{1}{2 \sqrt{t}}\frac{\sqrt[3]{2}}{2} .
$$