One. (15 points) Let $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, and $a_{n+1}=2 S_{n}+\frac{1}{2}\left(n \in \mathbf{Z}_{+}\right)$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$.
(2) Insert $n$ real numbers between $a_{n}$ and $a_{n+1}$ such that these $n+2$ numbers form an arithmetic sequence with a common difference of $d_{n}$. Let the sum of the first $n$ terms of the sequence $\left\{\frac{1}{d_{n}}\right\}$ be $T_{n}$. Prove:
$$
T_{n}<\frac{15}{4}\left(n \in \mathbf{Z}_{+}\right) .
$$
(1) Since $a_{n+1}=2 S_{n}+\frac{1}{2}(n \geqslant 1)$, therefore, $a_{n}=2 S_{n-1}+\frac{1}{2}(n \geqslant 2)$.
Subtracting the two equations gives
$$
\begin{array}{l}
a_{n+1}-a_{n}=2\left(S_{n}-S_{n-1}\right)=2 a_{n} \\
\Rightarrow a_{n+1}=3 a_{n}(n \geqslant 2) .
\end{array}
$$

Since $\left\{a_{n}\right\}$ is a geometric sequence, for any $n \in \mathbf{Z}_{+}$, we have $a_{n+1}=3 a_{n}$.
Also, from $a_{2}=2 a_{1}+\frac{1}{2}$, we get
$$
3 a_{1}=2 a_{1}+\frac{1}{2} \Rightarrow a_{1}=\frac{1}{2} \text {. }
$$

Thus, $a_{n}=\frac{1}{2} \times 3^{n-1}$.
(2) From the given condition,
$$
\begin{array}{l}
a_{n+1}=a_{n}+(n+1) d_{n} \\
\Rightarrow \frac{1}{d_{n}}=\frac{n+1}{a_{n+1}-a_{n}}=\frac{n+1}{3^{n-1}} .
\end{array}
$$

Then $T_{n}=2+\frac{3}{3}+\frac{4}{3^{2}}+\cdots+\frac{n+1}{3^{n-1}}$.
Thus, $\frac{1}{3} T_{n}=\frac{2}{3}+\frac{3}{3^{2}}+\cdots+\frac{n}{3^{n-1}}+\frac{n+1}{3^{n}}$.
Subtracting the above two equations gives
$$
\begin{array}{l}
\frac{2}{3} T_{n}=2+\frac{1}{3}+\frac{1}{3^{2}}+\cdots+\frac{1}{3^{n-1}}-\frac{n+1}{3^{n}} \\
=2+\frac{\frac{1}{3}\left(1-\frac{1}{3^{n-1}}\right)}{1-\frac{1}{3}}-\frac{n+1}{3^{n}} \\
=\frac{5}{2}-\frac{2 n+5}{2 \times 3^{n}} .
\end{array}
$$

Therefore, $T_{n}=\frac{15}{4}-\frac{2 n+5}{4 \times 3^{n-1}}<\frac{15}{4}$.