Proving that a Sequence Approaches Infinity Using Lagrange's Theorem Let $f(x)$ be a differentiable function in $(0,\infty)$, so that $f'(x)&gt;x$ for every $x&gt;0$.

$(a_n)$  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 $f'(x)&gt;x$ for every $x&gt;0$, we also know that

$$
a_{n+1}&gt;c_n &gt;a_n 
$$

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.
For all $x$ you have $f'(x)-x &gt;0$.

Then
$$0&lt; \int_{a_n}^{a_n+1} [f'(x) -x] \mathrm dx = f(a_n+1)-f(a_n)-\frac{1}{2}((a_n+1)^2-a_n^2)=a_{n+1}-a_n-\frac{1}{2}$$
From this follows that
$$a_{n+1}&gt;a_n + \frac{1}{2}$$
and this is enough to conclude that $a_n$ is unbounded.
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 

$$
\sup{a_n} = \lim_{n \to \infty}a_n =L &gt;0
$$

$f$ 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 $t \in (L,L+1)$,

$$
f'(t) = \frac{f(L+1)-f(L)}{L+1-L} = f(L+1)-f(L) &gt; t &gt; L
$$

But from $(\star)$, we get 

$$
L &gt; L
$$

Which is a contradiction.