14. (15 points) Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=2 p, a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{p^{2}}{a_{n}}\right), \\
b_{n}=\frac{a_{n}+p}{a_{n}-p}\left(n \in \mathbf{N}_{+}, p>0\right) .
\end{array}
$$
(1) Find the general term of the sequence $\left\{b_{n}\right\}$;
(2) Prove: $\frac{a_{n}-p}{a_{n+1}-p}=3^{2^{n-1}}+1$;
(3) Let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. When $n \geqslant 2$, is the relationship between $S_{n}$ and $\left(n+\frac{23}{18}\right) p$ determined? Please explain your reasoning.
14. (1) From the given, we have
$$
\begin{array}{l}
b_{n+1}=\frac{a_{n+1}+p}{a_{n+1}-p}=\frac{a_{n}+\frac{p^{2}}{a_{n}}+2 p}{a_{n}+\frac{p^{2}}{a_{n}}-2 p} \\
=\frac{\left(a_{n}+p\right)^{2}}{\left(a_{n}-p\right)^{2}}=b_{n}^{2}>0 .
\end{array}
$$

Then $\lg b_{n+1}=2 \lg b_{n}$
$$
\Rightarrow \lg b_{n}=2^{n-1} \lg 3 \Rightarrow b_{n}=3^{2^{n-1}} \text {. }
$$
(2) From the given, we have
$$
\begin{array}{l}
a_{n}=\frac{b_{n}+1}{b_{n}-1} \cdot p=p+\frac{2 p}{3^{2^{n-1}}-1} . \\
\text { Then } \frac{a_{n}-p}{a_{n+1}-p}=\frac{3^{2^{n}}-1}{3^{2^{n-1}}-1}=3^{2^{n-1}}+1 .
\end{array}
$$
(3) When $n \geqslant 2$,
$$
a_{n+1}-p=\frac{a_{n}-p}{3^{2^{n-1}}+1} \leqslant \frac{1}{10}\left(a_{n}-p\right) \text {. }
$$

Then $S_{2}=a_{1}+a_{2}=2 p+\frac{5}{4} p<\left(2+\frac{23}{18}\right) p$.
When $n \geqslant 3$,
$$
a_{k}-p \leqslant \frac{1}{10}\left(a_{k-1}-p\right)(k=3,4, \cdots, n) \text {. }
$$

Then $S_{n}-a_{1}-a_{2}-(n-2) p$
$$
\begin{aligned}
\leqslant & \frac{1}{10}\left[S_{n-1}-a_{1}-(n-2) p\right] \\
\Rightarrow & 10 S_{n}-\frac{65}{2} p-10(n-2) p \\
& \leqslant S_{n}-a_{n}-2 p-(n-2) p \\
\Rightarrow & S_{n} \leqslant\left[(n-2)+\frac{61}{18}-\frac{3^{2^{n-1}}+1}{9\left(3^{2^{n-1}}-1\right)}\right] p \\
& <\left(n+\frac{23}{18}\right) p .
\end{aligned}
$$

In summary, when $n \geqslant 2$, $S_{n}<\left(n+\frac{23}{18}\right) p$.