9. (This question is worth 16 points) 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 $\left\{a_{n}\right\}$;
(2) If there exists a real number $c$ such that $\sum_{i=1}^{k} \frac{1}{a_{i}}<c$ for all positive integers $k$, determine the range of values for $c$.
Solution: (1) In $a_{m+n}=a_{m}+a_{n}+2 m n$, let $m=1$ to obtain the recursive formula for $\left\{a_{n}\right\}$:
$$
a_{n+1}=a_{1}+a_{n}+2 n=a_{n}+(3+2 n) \text {. }
$$

Therefore, the general term formula for $\left\{a_{n}\right\}$ is:
$$
a_{n}=a_{1}+\sum_{k=1}^{n-1}(3+2 k)=3+\frac{[5+(2 n+1)](n+1)}{2}=n(n+2) .8 \text { points }
$$
(Indeed, for this sequence $\left\{a_{n}\right\}, a_{1}=1 \times 3=3$, and
$$
\begin{array}{l}
a_{m+n}=(m+n)(m+n+2)=(m+n)^{2}=2(m+n)=\left(m^{2}+2 m\right)+2\left(n^{2}+2 n\right)+2 m n \\
=a_{m}+a_{n}+2 m n .
\end{array}
$$

Thus, $a_{n}=n(n+2)$ is the general term formula for the sequence $\left\{a_{n}\right\}$.
(2) Note that: $\frac{1}{a_{n}}=\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$, so
$$
\sum_{n=1}^{k} \frac{1}{a_{n}}=\sum_{n=1}^{k} \frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{k+1}-\frac{1}{k+2}\right)=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{k+1}+\frac{1}{k+2}\right) \text {. }
$$

Therefore, $\sum_{n=1}^{k} \frac{1}{a_{n}}<\frac{3}{4}$, and $\sum_{n=1}^{k} \frac{1}{a_{n}} \rightarrow \frac{3}{4}(k \rightarrow \infty)$, so the range of $c$ is $c \in\left[\frac{3}{4},+\infty\right) .16$ points