How to prove a sequence is diverging to infinity using derivative properties?

Question

Let $f(x)$ is continues and has derivative in >$(0,\infty)$ and satisfy $f'(x)>x$ for every $x \in >(0,\infty)$ and let $(a_n)$ be a sequence such that $a_1 = 1$ and for each $n>1$, $a_{n+1} = f(a_n + 1)-f(a_n)$.

Prove that: $lim_{n \rightarrow \infty} a_n = \infty$

I tried to use Lagrange theorem: $f$ satisfy the conditions in every $[x,x+1]$ for $x>0$, so: $\frac{f(x+1)-f(x)}{x+1-x}>f'(c)>x$ so I get: $a_{n+1}=f(a_{n+1})-f(a_n)>a_n$ So $(a_n)$ is monotonically increasing. But how do I prove it is not bounded?

Thanks!

Answer 1

Since $(a_n)$ is monotone, it has a limit $L$. If $L<\infty$, continuity of $f$ and the relation $a_{n+1} = f(a_n+1) - f(a_n)$ implies $L = f(L+1) - f(L)$. But by the mean value theorem, $f(L+1) - f(L) = f'(c)$ for some $c\in(L,L+1)$, and in particular $f(L+1) - f(L) > L$, a contradiction. Hence $L=\infty$.

Answer 2

I thought of the following: from $f'(x)>x$ we can prove that for each $n \in \mathbb N$, $f(x)>nx$ for every $x>n$ (otherwise $f'(x) \leq n$ for some $x$ contradicting $f'(x)>x,\forall x \in \mathbb R$).

So suppose $(a_n)$ is bounded, then it has a limit $L$ and since it is monotonically increasing, $L > a_n, \forall n \in \mathbb N$. $f$ is increasing too so $\forall n \in \mathbb N, f(L)>f(a_n)$ and denote $f(L)$ by $K$.

So, $ \forall n \in \mathbb N$, $a_{n+1} > n(a_{n+1} + 1) - K \Rightarrow lim_{n \rightarrow \infty}(a_{n+1}) > lim_{n \rightarrow \infty}(n(a_{n+1} + 1) - K) = lim_{n \rightarrow \infty}(na_{n+1} + n - K) > lim_{n \rightarrow \infty}( n - K) = (lim_{n \rightarrow \infty}n) - K = \infty$ and from sequence comparison criteria we get, $(lim_{n \rightarrow \infty}a_n = \infty$.

What do you think? looks OK?