9. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3$, and for any positive integers $m, n$, it holds that $a_{m+n}=a_{m}+a_{n}+2 m n$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If a real number $c$ makes $\sum_{i=1}^{k} \frac{1}{a_{i}}<c$ hold for all positive integers $k$, find the range of values for $c$.
(1) Let $m=1 \Rightarrow a_{n+1}=a_{n}+a_{1}+2 n \Rightarrow a_{n+1}-a_{n}=2 n+3$,
\[
\left\{\begin{array}{l}
a_{n}-a_{n-1}=2 n+1, \\
a_{n-1}-a_{n-2}=2 n-1, \\
\cdots \cdots \\
a_{2}-a_{1}=5, \\
a_{1}=3
\end{array} \Rightarrow a_{n}=(n+1)^{2}-1=n^{2}+2 n, n \in \mathbf{N}^{*}\right.
\]
(2) $\frac{1}{a_{i}}=\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$, thus $\sum_{i=1}^{k} \frac{1}{a_{i}}=\frac{1}{2} \sum_{i=1}^{k}\left(\frac{1}{n}-\frac{1}{n+2}\right)$ $=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)=\frac{3}{4}-\frac{2 n+3}{2(n+1)(n+2)}<\frac{3}{4}$.
Since $n \rightarrow \infty$ implies $\frac{2 n+3}{2(n+1)(n+2)} \rightarrow 0 \Rightarrow \sum_{i=1}^{k} \frac{1}{a_{i}} \rightarrow \frac{3}{4} \Rightarrow c \geqslant \frac{3}{4}$,
Therefore, the range of $c$ is $\left[\frac{3}{4},+\infty\right)$.