smc

Because we start with $x_1=-1$, and the terms $x_k$ alternate between $1$ and $-1$, there is either one more $-1$ than the number of $1$s (when $n$ is odd), or there are an equal number of $1$s and $-1$s (when $n$ is even). In the former case, $f(n)=\frac{-1}{n}$, and, in the latter case, $f(n)=0$. This is only consistent with answer $\boxed{\{0,-\frac{1}{n}\}}$.