3
$\begingroup$

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.

2 Answers 2

4

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, $\displaystyle\lim_{n\rightarrow\infty} \frac{f(n)}{n}=L$

  • 0
    Yes, I just found this...and consequently, http://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem Thank you.2012-10-11
4

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 i.e. for $n\geqslant n_{\varepsilon}$ $L-\varepsilon Adding these inequalities, we have $(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 $\lim\limits_{n\rightarrow\infty}\dfrac{f_n}{n}=L$