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Here's the problem in full. I've stared at it for hours and can't get anywhere, so a hint would be nice.

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 $L$.

This problem is marked with the word "Cauchy", so I'm guessing Cauchy sequences or something related to them will be necessary, but I can't figure out what.

Edit: It appears as if the Stolz-Cesaro Theorem makes this quite easy. Unfortunately though, we have not discussed this, so I won't be able to use this.

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To make the indices work nicely, we use a completely unnecessary little trick. Our function $f$ is presumably defined over the positive integers. It is convenient to extend the definition to non-negative integers, by letting $f(0)=0$. Note that $$f(n)=f(n)-f(0)=(f(1)-f(0))+(f(2)-f(1))+\cdots +(f(n)-f(n-1)).$$ Letting $g(n)=f(n)-f(n-1)$, and dividing by $n$, we get $$\frac{f(n)}{n}=\frac{g(1)+g(2)+\cdots +g(n)}{n},\tag{1}$$ We want to show that the right-hand side of (1) has limit $L$. We have $$f(n)-nL=(g(1)-L)+(g(2)-L)+\cdots+(g(n)-L),$$ so by the Triangle Inequality $$|f(n)-nL|\le \sum_1^n |g(i)-L|.\tag{2}$$

Let $\epsilon\gt 0$ be given. There is an $M$ such that if $k\gt M$ then $|g(k)-L|\lt \epsilon/2$. Let $B=\sum_1^M |g(i)-L|$. If $n\gt M$, then the right side of (2) is less than $$B+(n-M)\frac{\epsilon}{2}.$$ Dividing by $n$, we find that for $n$ beyond $M$ we have $$\left|\frac{f(n)}{n}-L\right|\lt \frac{B}{n}+\frac{n-M}{n}\cdot\frac{\epsilon}{2}.\tag{3}$$

Finally, by choosing $n$ large enough and beyond $M$, we can make the right-hand side of (3) less than $\epsilon$. All we need to do is to make $\frac{B}{n}\lt \frac{\epsilon}{2}$.

Remark: We gave a quite detailed proof, as a sample of how to handle the epsilons. The details may hide the basic simplicity of the idea. After we use inverse telescoping to express $f(n)-nL$ as a sum of the $g(i)-L$, the rest is geometrically quite natural.

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Hint : Check the cesaro theorem regarding arithmetical mean of a sequence, when it converges towards a limit L

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  • $\begingroup$ While this is a relatively easy solution, I can't use this theorem because it hasn't been discussed yet. $\endgroup$ – wlyles Oct 16 '14 at 17:12
  • $\begingroup$ I see, then use epsilon and sum the relation with the right index. $\endgroup$ – mvggz Oct 16 '14 at 17:33
  • $\begingroup$ If you define : $U_n = f(n+1) - f(n)$ , then: $\endgroup$ – mvggz Oct 16 '14 at 17:35
  • $\begingroup$ For any ϵ>0, there is a $N_ϵ$ such as: n> $N_ϵ$ , |$U_n$ -L| < ϵ sum this from $N_ϵ$ + 1 to n-1 , you will get $\frac{f(n)}{n}$ and a residual term that ->0 (a constant A/n) You can conclude with this $\endgroup$ – mvggz Oct 16 '14 at 17:38
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Hint: try to prove the case $L=0$ first.

If $N$ is such as if $n>n$ then $|f(n+1) - f(n)| < \epsilon$, then
$$\frac 1n\sum_{k=N}^n |f(k+1) - f(k)| < \epsilon$$

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Note that $f(n+1)-f(n) = \dfrac{f(n+1)-f(n)}{(n+1)-n}$. Then, by Stolz-Cesàro applied to the sequence $\dfrac{f(n)}{n}$ we have:

$$\lim_{n \to \infty} \dfrac{f(n+1) - f(n)}{n+1 - n} = L \Rightarrow \lim_{n \to \infty} \dfrac{f(n)}{n}= L$$

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  • $\begingroup$ While this is a relatively easy solution, I can't use this theorem because it hasn't been discussed yet. $\endgroup$ – wlyles Oct 16 '14 at 17:15

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