jmc

Note that $n+1$ and $n$ will never share any common factors except for $1$, because they are consecutive integers. Therefore, $n/(n+1)$ is already simplified, for all positive integers $n$.

Since $1\le n\le 100$, it follows that $2\le n+1\le 101$. Recall that a simplified fraction has a repeating decimal representation if and only if its denominator is divisible by a prime other than 2 and 5. The numbers between 2 and 101 which are divisible only by 2 and 5 comprise the set $\{2,4,5,8,10,16,20,25,32,40,50,64,80,100\}$. Therefore, there are $14$ terminating decimals and $100-14=\boxed{86}$ repeating decimals.