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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.

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2 Answers 2

up vote 4 down vote accepted

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

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Yes, I just found this...and consequently, en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem Thank you. –  kaiserphellos Oct 11 '12 at 9:20
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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 $\lim\limits_{n\rightarrow\infty}\dfrac{f_n}{n}=L$

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