China Mathematical Competition (Shaanxi)

Since $a_1,a_2,\dots ,a_{n-1}$ all have exactly two possible values, so $T_n=2^{n-1}$. Meanwhile, the frequency of $a_i=1$ is the same as that of $a_i=0$ for $1\le i\le n-1$, and $a_n=1$. Then $$\begin{array}{rl}{S}_n& =\frac{1}{2}\times 2^{n-1}\times \left(\frac{1}{10}+\frac{1}{10^2}+\cdots +\frac{1}{10^{n-1}}\right)+2^{n-1}\times \frac{1}{10^n}\\ & =2^{n-1}\times \frac{1}{18}\times \left(1-\frac{1}{10^{n-1}}\right)+2^{n-1}\times \frac{1}{10^n}.\end{array}$$

So $$\underset{n\to \infty }{\lim }\frac{S_n}{T_n}=\underset{n\to \infty }{\lim }\left[\frac{1}{18}\left(1-\frac{1}{10^{n-1}}\right)+\frac{1}{10^n}\right]=\frac{1}{18}.$$