hayyyyyyy....

This is in the wrong forum, I will move it.

Let $U_n(x)$ be the sequence such that $U_0(x)=0 , U_{n+1}=\int_0^x \frac {1}{1+u_n (t)}dt.$. It's easy to see that $0=U_0(x) \leq... \leq U_{2n}(x) \leq f_{2n+1}(x) \leq U_{2n+1}(x) \leq ... \leq U_1$ and $ U_0(x) \leq... \leq U_{2n+2}(x) \leq f_{2n+2}(x) \leq U_{2n+1}(x) \leq ... \leq U_1 $ Manifestly $U_1-U_0=x$ and $0 \leq U_{2m+1}(x)-U_{2n}(x) \leq \int_0^x (U_{2n-1}(t)-U_{2m}(t))dt$ so : $U_1(x)-U_2(x) \leq \frac{x^2}2 , ..., U_{2n+1}-U_{2n} \leq \frac {x^{2n+1}}{(2n+1)!} , U_{2n+1}-U_{2n+2 } \leq \frac {x^{2n+1}}{(2n+2)!}, ...$ Now it's clear that $\{U_{2n+1}(x)\}$ and $\{U_{2n+1}(x)\}$ converge to the same limit $\phi (x)$, so $f_n(x)$ converges to $\phi (x)$: $\phi (x) = \int_0^x \frac {1}{1+\phi (t)}dt \Longrightarrow \phi '(x) = \frac {1}{1+\phi (x)} \Longrightarrow \phi (x) = \sqrt {1+2x}-1 $