A sequence of functions defined by a Riemann integral 
Consider a sequence of functions $(f_n)_{n\in \mathbb{N}}$, where $f_n:[0,1]\to [0,\infty)$, $f_0$ is an arbitrary continuous function and $$f_{n+1}(x)=\int\limits_0^x\frac{1}{1+f_n(t)}dt, \forall x\in [0,1], n\in \mathbb{N}.$$
  Prove that $\forall x\in [0,1]$ the sequence $(f_n(x))_{n\in \mathbb{N}}$ converges and find its limit. 

Let's note that $f_n$ is continuous $\forall n\ge 0$. Assume that $f:[0,1]\to \mathbb{R}$ is a function such that $f(x)=\int\limits_0^x \frac{dt}{1+f(t)}$, $\forall x\in [0,1]$, then $f'(x)=\frac{1}{1+f(x)}, \forall x\in[0,1] \implies f(x)+\frac{f^2(x)}{2}-x=0, \forall x\in [0,1] $
$\implies f(x)=\sqrt{1+2x}-1, \forall x\in [0,1]$.
My idea was to show that $\lim \limits_{n\to \infty}f_n(x)=f(x), \forall x\in [0,1]$, but I couldn't do this. 
 A: One can remark that :
$$ f_{n+2}(x) - f_{n+1}(x) = \int_0^x \frac{f_n(t) - f_{n+1}(t)}{(1+f_n(t))(1+f_{n+1}(t))} dt $$
$$ ||f_{n+2} - f_{n+1}||_{\infty} \leq ||f_{n+1}-f_n||_{\infty}\int_0^1 \frac{1}{(1+f_n(t))(1+f_{n+1}(t))} dt $$
For $n\geq 1$, since $f_{n-1}$ is non-negative, for all $x \in [0,1]$, $f_n(x) \leq 1$ therefore $\displaystyle\frac{1}{1+f_n(x)} \geq \frac{1}{2}$.
This yields for any $n\geq 2$ :
$$f_n(t) = \int_0^t \frac{1}{1+f_{n-1}(s)}ds \geq \frac{t}{2}$$
$$(1+f_n(t))(1+f_{n+1}(t)) \geq \frac{(t+2)^2}{4}$$
$$\frac{1}{(1+f_n(t))(1+f_{n+1}(t))} \leq \frac{4}{(t+2)^2}$$
Integrating on $[0,1]$ yields :
$$\int_0^1 \frac{1}{(1+f_n(t))(1+f_{n+1}(t))} \leq \left[-\frac{4}{t+2}\right]_0^1 = \frac{2}{3}$$
Thus we have :
$$ ||f_{n+2} - f_{n+1}||_{\infty} \leq \frac{2}{3}||f_{n+1}-f_n||_{\infty}$$
An immediate induction yields, that there exists $C \geq 0$ such that
$$
||f_{n+1} - f_{n} || \leq C\left(\frac{2}{3}\right)^n
$$
Now for $m \geq n$ we have : $$||f_{m} - f_{n}||_{\infty}\leq \sum_{i = 0}^{m-n} ||f_{n+i+1} - f_{n+i}||_{\infty}\leq ||f_{n+1} - f_{n}||_{\infty}\sum_{i = 0}^{m-n} (\frac{2}{3})^{i} \leq C'(\frac{2}{3})^n$$
So $(f_n)$ is a Cauchy sequence in the complete space $\mathcal{C}([0,1])$ equipped with the norm $||.||_\infty$, therefore it converges uniformly to a continuous and non-negative function $f\in \mathcal{C}([0,1])$.
We have for any $x\in [0,1]$, due to the non-negativity of $f$ and the dominated convergence theorem :
$$\int_0^x \frac{1}{1+f_n(t)} dt \longrightarrow \int_0^x \frac{1}{1+f(t)} dt$$
And by uniqueness of the limit, we have :
$$f(x) = \int_0^x \frac{1}{1+f(t)} dt$$
From there you did well !
