Proving that a Sequence Approaches Infinity Using Lagrange's Theorem Let $f(x)$ be a differentiable function in $(0,\infty)$, so that $f'(x)>x$ for every $x>0$.
$(a_n)$  is the sequence defined recursively as
$$
a_1 = 1\\
a_{n+1}=f(1+a_n)-f(a_n)
$$
Q. Prove that 
$$
\lim_{n\to\infty}a_n=\infty
$$
Using Lagrange's theorem, it's easy to show that for each interval $I_n=[a_n,a_n+1]$, there's some $c_n \in (a_n,a_n+1)$ so that
$$
f'(c_n) = \frac{f(1+a_n)-f(a_n)}{1+a_n-a_n} = f(1+a_n)-f(a_n)=a_{n+1}
$$
And by the the fact that $f'(x)>x$ for every $x>0$, we also know that
$$
a_{n+1}>c_n >a_n 
$$
So, we've proved that $(a_n)$ is increasing and positve, but I cannot develop this further to prove that it is not bounded (to conclude it approaces infinity).
How can I prove this last step?
Note: I'm looking primarily for a solution without integrals. The question should be solveable with differential calculus alone - derivative, continuity, and limit of sequence.
 A: For all $x$ you have $f'(x)-x >0$.
Then
$$0< \int_{a_n}^{a_n+1} [f'(x) -x] \mathrm dx = f(a_n+1)-f(a_n)-\frac{1}{2}((a_n+1)^2-a_n^2)=a_{n+1}-a_n-\frac{1}{2}$$
From this follows that
$$a_{n+1}>a_n + \frac{1}{2}$$
and this is enough to conclude that $a_n$ is unbounded.
A: Suppose $\{a_n\}$ is bounded. Since you have shown that $a_n$ is monotone increasing, then $a_n\rightarrow a$ for some $a>0$. Hence if $f$ is continuous then you have
\begin{align}
f'(a) = a
\end{align}
for some $a$...contradiction.
Edit: Otherwise, observe you have
\begin{align}
a=f(1+a)-f(a) = f'(c)> c
\end{align}
for some $c \in (a, a+1)$. Hence a contradiction.
A: After fiddling with this question for a while, I've arrived at the following solution, which does not require any further assumptions.
Suppose that $(a_n)$ is bounded. We've shown that it is monotone increasing and positive, so there exists some $L\in \mathbb{R}$ so that 
$$
\sup{a_n} = \lim_{n \to \infty}a_n =L >0
$$
$f$ is differentiable in $\mathbb{R}$, and in particular continuous in $L$ and $L+1$, so
$$
\lim_{n \to \infty}{f(a_n+1)-f(a_n)} = f(L+1)-f(L)
$$
From the definition of $a_{n+1}$, we get
$$
L=\lim_{n \to \infty}{a_{n}} = \lim_{n \to \infty}{a_{n+1}} = \lim_{n \to \infty}{f(1+a_n)-f(a_n)} = f(L+1)-f(L)
$$
So
$$
(\star) \quad L = f(L+1)-f(L)
$$
Now, using Lagrange's theorem again with the interval $[L,L+1]$, we get that for some $t \in (L,L+1)$,
$$
f'(t) = \frac{f(L+1)-f(L)}{L+1-L} = f(L+1)-f(L) > t > L
$$
But from $(\star)$, we get 
$$
L > L
$$
Which is a contradiction.
