Exercise on Limits of Functions I have an exercise that states:
Suppose that $f:\mathbb{N}\rightarrow\mathbb{R}$. If $\lim_{n\rightarrow\infty}f(n+1)-f(n)=L$, prove that $\lim_{n\rightarrow\infty}f(n)/n$ exists and equals $L$.
I have wracked my brain over this exercise for about 5 hours now and really have not gotten ANYWHERE with it...does anyone have a suggestion on how begin proving this?
Thank you.
 A: Using Cesàro mean
http://en.wikipedia.org/wiki/Ces%C3%A0ro_mean
to obtain the solution.
Let $a_{n+1}=f(n+1)-f(n) (n\in \mathbb{N})$. Then we have
$$
f(n)=[f(n)-f(n-1)]+[f(n-1)-f(n-2)]+\ldots+[f(2)-f(1)]+[f(1)-f(0)]+f(0)=a_n+a_{n-1}+\ldots+a_1+f(0)
$$
and so
$$
\frac{f(n)}{n}=\frac{\displaystyle\sum_{i=1}^{n}a_i}{n}+\frac{f(0)}{n}.
$$
Since $\lim_{n\rightarrow\infty} a_n=L$ then by the Cesaro's theorem 
$$
\lim_{n\rightarrow\infty}\frac{\displaystyle\sum_{i=1}^{n}a_i}{n}=L.
$$
Hence, $\displaystyle\lim_{n\rightarrow\infty} \frac{f(n)}{n}=L$
A: If exists $\lim\limits_{n\rightarrow\infty}(f_{n+1}-f_{n})=L,$ then by definition
$$(\forall \varepsilon>0)\quad (\exists n_{\varepsilon}\in \mathbb{N}):\quad (\forall n\geqslant n_{\varepsilon})\\L-\varepsilon<f_{n+1}-f_{n}<L+\varepsilon,$$ i.e. for $n\geqslant n_{\varepsilon}$
$$L-\varepsilon<f_{n_{\varepsilon}+1}-f_{n_{\varepsilon}}<L+\varepsilon\\
L-\varepsilon<f_{n_{\varepsilon}+2}-f_{n_{\varepsilon}+1}<L+\varepsilon\\
\vdots\\L-\varepsilon<f_{n}-f_{n-1}<L+\varepsilon.$$ Adding these inequalities, we have
$$(n-n_{\varepsilon})(L-\varepsilon)<f_{n}-f_{n_{\varepsilon}}<(n-n_{\varepsilon})(L+\varepsilon),$$
$$\left(1-\dfrac{n_{\varepsilon}}{n}\right)(L-\varepsilon)<\dfrac{f_{n}-f_{n_{\varepsilon}}}{n}<\left(1-\dfrac{n_{\varepsilon}}{n}\right)(L+\varepsilon),\\
\left(1-\dfrac{n_{\varepsilon}}{n}\right)(L-\varepsilon)+\dfrac{f_{n_{\varepsilon}}}{n}<\dfrac{f_{n}}{n}<\left(1-\dfrac{n_{\varepsilon}}{n}\right)(L+\varepsilon)+\dfrac{f_{n_{\varepsilon}}}{n}.
$$
Therefore, (omitting some technical details) we may conclude that exists $\lim\limits_{n\rightarrow\infty}\dfrac{f_n}{n}=L$
