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

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.

• if $f(n)=\sin(n\pi/2)$, the series converges but the sequence does not. You need to add some conditions to the question. – Mefitico May 25 '18 at 20:19
• @KennyLau : Fair enough, editing comment. – Mefitico May 25 '18 at 20:21
• But then of course, the question is incorrect if you do not stipulate that the limits exist. – Kenny Lau May 25 '18 at 20:21
• We never need the continuity of $f$. In fact, we just care about the value of $f(x)$ for $x\in\mathbb{N}$. It is better to re-phase your question using the word "sequence of real nubmers". – Danny Pak-Keung Chan May 25 '18 at 21:51

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