If ‎$‎f$ ‎is a ‎concave function ‎then ‎$‎f$ ‎is ‎increasing‎ My question is how to prove the following assertion: 

If $f:\mathbb{R}\to\mathbb{R}$ is a concave function such that $\lim\limits_{x\to\infty}(f(x)-f(x-1))=0$ then $f$ increasing.  

I know that if ‎$‎‎f$ ‎is ‎differentiable ‎then the ‎mentioned ‎hypothesis ($f'$ is decreasing) and using M.V.T ‎imply ‎that firstly ‎‎$\lim\limits_{x\to\infty}f'‎(x)=0$ also ‎‎$‎f'‎\geq0$ and $‎‎f$‎ ‎is ‎increasing. ‎but ‎here ‎we ‎do not ‎have ‎the differentiability.
‎Anyone ‎can ‎help ‎me. Thanks.‎

Hint. Since $f$ is concave it follows that for any $x_0\in \mathbb{R}$ the function
$$x\to R(x,x_0):=\frac{f(x)-f(x_0)}{x-x_0}$$
is decreasing in $\mathbb{R}\setminus\{x_0\}$. Here it is a reference for the convex function $-f$.
Now assume that $f$ is not increasing. Then there are $x_0,x_1$ such that $x_0<x_1$ and $f(x_1)<f(x_0)$. It follows that
$$0>R(x_1,x_0)\geq R(x_0+k+1,x_0)\geq R(x_0+k+1,x_0+k)$$
where $k$ is any positive integer such that $x_0+k>x_1$.
Can you take it from here?
