1. Given the sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=\log _{3}\left(1+\frac{2}{n^{2}+3 n}\right)$, then $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=$
1. 1 .

From $a_{n}=\log _{3} \frac{n^{2}+3 n+2}{n^{2}+3 n}=\log _{3} \frac{(n+1)(n+2)}{n(n+3)}$ we get
$$
\begin{aligned}
a_{1}+a_{2}+\cdots+a_{n} & =\log _{3} \frac{2 \times 3}{1 \times 4}+\log _{3} \frac{3 \times 4}{2 \times 5}+\cdots+\log _{3} \frac{(n+1)(n+2)}{n(n+3)} \\
& =\log _{3}\left[\frac{2 \times 3}{1 \times 4} \cdot \frac{3 \times 4}{2 \times 5} \cdots \cdots \frac{(n+1)(n+2)}{n(n+3)}\right] \\
& =\log _{3} \frac{3(n+1)}{n+3} .
\end{aligned}
$$

Therefore
$$
\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\cdots+a_{n}\right)=\lim _{n \rightarrow \infty} \log _{3} \frac{3(n+1)}{n+3}=\log _{3}\left(\lim _{n \rightarrow \infty} \frac{3(n+1)}{n+3}\right)=1
$$