Given that $f_1(x) = \frac{x}{x + 1}$ and $f_{n + 1} (x) = f_1 (f_n(x))$ for $n \ge 1$, then find $f_{2014}(x)$.
Just by writing the first few terms the pattern becomes clear. In general \(f_n(x)=\frac{x}{nx+1}\)
