15. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}>0$, and
$$
a_{n+1}=\sqrt{\frac{3+a_{n}}{2}} \text {. }
$$
(1) Find the range of $a_{1}$ such that $a_{n+1}>a_{n}$

for any positive integer $n$;
(2) If $a_{1}=4$, let $b_{n}=\left|a_{n+1}-a_{n}\right|(n=1$, $2, \cdots)$, and let $S_{n}$ denote the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$. Prove: $S_{n}<\frac{5}{2}$.
15. (1) Research
$$
\begin{array}{l}
a_{n+1}-a_{n}=\sqrt{\frac{3+a_{n}}{2}}-\sqrt{\frac{3+a_{n-1}}{2}} \\
=\frac{a_{n}-a_{n-1}}{2\left(\sqrt{\frac{3+a_{n}}{2}}+\sqrt{\frac{3+a_{n-1}}{2}}\right)}(n \geqslant 2) .
\end{array}
$$

Notice that $\sqrt{\frac{3+a_{n}}{2}}+\sqrt{\frac{3+a_{n-1}}{2}}>0$.
Therefore, $a_{n+1}-a_{n}, a_{n}-a_{n-1}, \cdots, a_{2}-a_{1}$ have the same sign.

To ensure that $a_{n+1}>a_{n}$ for any positive integer, it is sufficient that $a_{2}-a_{1}>0$.
From $\sqrt{\frac{3+a_{1}}{2}}-a_{1}>0$, solving gives $0\frac{3}{2}$, then $a_{n+1}\frac{3}{2}$. Hence $S_{n}<4-\frac{3}{2}=\frac{5}{2}$.