smc

Note that$$0.{\overline{133}}_n=\frac{n^2+3n+3}{n^3-1},$$by geometric series. Thus, we're aiming to find the value of$$\prod _{k=4}^{99}\frac{k^2+3k+3}{k^3-1}.$$Expanding the product out, this is equivalent to $$\frac{4^2+3(4)+3}{4^3-1}\cdot \frac{5^2+3(5)+3}{5^3-1}\cdot \frac{6^2+3(6)+3}{6^3-1}\cdot ...\cdot \frac{99^2+3(99)+3}{99^3-1}.$$Note that the numerator of the $a$th fraction and the denominator of the $a+1$th fraction for $1\le a\le 95$ cancel out to be $\frac{1}{a+3},$ since$$\frac{k^2+3k+3}{(k+1{)}^3-1}=\frac{k^2+3k+3}{k^3+3k^2+3k}=\frac{1}{k},$$by the binomial theorem on the the denominator of the aforementioned. Since this forms a telescoping series, our product is now equivalent to$$\frac{99^2+3(99)+3}{4^3-1}\cdot \prod _{k=4}^{98}\frac{1}{k},$$which, after simplification gives $\frac{6}{98!}\cdot \frac{10101}{63}=\frac{962}{98!},$ giving an answer of $\boxed{962}.$