Why is $\lim_{ n\rightarrow \infty} \frac{1}{n} \sum_{h < n} f(h) = \lim_{ n\rightarrow \infty} f(n) $.

Question

Given $f$ a continuous function

$$\lim_{ n\rightarrow \infty} \frac{1}{n} \sum_{h < n} f(h) = \lim_{ n\rightarrow \infty} f(n) $$

How could one prove this?

Edit: as rightly stated in the comments assuming that $\lim_{ n\rightarrow \infty} f(n)$ exists.

Answer 1

This is just a basic result known as Cesàro summation, a discussion of the proof can be found here

Answer 2

By the Stolz theorem:

$\lim\limits_{n\to \infty} \frac{1}{n} \sum_{k=0}^n f(k) = \lim\limits_{n\to \infty} \frac{\sum_{k=0}^n f(k) - \sum_{k=0}^{n-1} f(k)}{n-(n-1)} = \lim\limits_{n\to \infty} f(n) $

Answer 3

How to understand this if you have never heard of Cesaro: assuming that $ f $ has a limit, when $n$ becomes large most of the terms in the left-hand side are going to be very close to that limit, and the contribution of the first few terms becomes comparatively small. 

In fact, by choosing $ n $ large enough you can make the terms that bother you contribute as little as you like to the average.

With this in mind it becomes easy to write down a formal $\varepsilon$-based proof.