An algebra problem by Bithiah Koshy

Given that $f_1(x) = \dfrac x{x+1}$ and $f_{n+1} (x) = f_1(f_n(x))$ , then

$\begin{aligned} f_1 (x) & = \frac x{x+1} \\ f_2 (x) & = f_1 (f_1(x)) = f_1 \left(\frac x{x+1} \right) = \frac {\frac x{x+1}}{\frac x{x+1}+1} = \frac x{2x +1} \\ f_3 (x) & = \frac {\frac x{2x+1}}{\frac x{2x+1}+1} = \frac x{3x+1} \end{aligned}$

It would appear that $f_n(x) = \dfrac x{nx+1}$ . Let us prove the claim is true for $n \ge 1$ by proof by induction . The claim is true for $n=1$ . Assuming that it is true for $n$ , then

$f_{n+1}(x) = f_1 (f_n (x)) = \frac {\frac x{nx+1}}{\frac x{nx+1}+1} = \frac x{(n+1)x+1}$

The claim is also true for $n+1$ , therefore it is true for all $n \ge 1$ . Then $f_{2014}(x) = \boxed{\dfrac x{2014x+1}}$ .