14. The first term of the sequence $\left\{a_{n}\right\}$ is 2, the sum of the first $n$ terms is $S_{n}$, and $\left\{S_{n}\right\}$ is a geometric sequence with a common ratio of $\frac{1}{3}$.
(1) Find the general formula for the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=a_{n} S_{n}$, find the sum of all terms of the sequence $\left\{S_{n}\right\}$.
14. Solution: (1) $S_{1}=a_{1}=2, S_{n}=2\left(\frac{1}{3}\right)^{n-1}$,
$$
a_{n}=S_{n}-S_{n-1}=2\left(\frac{1}{3}\right)^{n-1}-2\left(\frac{1}{3}\right)^{n-2}=2\left(\frac{1}{3}\right)^{n-1}(1-3)=-4\left(\frac{1}{3}\right)^{n-1} \text {. }
$$

So $a_{n}=\left\{\begin{array}{ll}2 & (n=1) \\ -4\left(\frac{1}{3}\right)^{n-1} & (n \geqslant 2)\end{array}\right.$.
(2) $b_{n}=a_{n} S_{n}=-4\left(\frac{1}{3}\right)^{n} \cdot 2\left(\frac{1}{3}\right)^{n-1}(n \geqslant 2)=-8\left(\frac{1}{3}\right)^{2(n-1)}$.

So $\lim _{n \rightarrow \infty} \sum_{i=1}^{n} b_{i}=a_{1} S_{1}+\lim _{n \rightarrow} \sum_{i=2}^{n} b_{i}=4+\frac{-8\left(\frac{1}{3}\right)^{2}}{1-\frac{1}{9}}=4-1=3$.