What does this notation represent? $\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^{n}a_n$ (Cesaro Means)

Question

My question is as the title, what does this represent?$$\lim_{n\to \infty}\frac{1}{n}\sum_{i=1}^{n}a_i$$ It looks like a series, but it is not as $\frac{1}{n}$ should not be there. Is this just a sequence that looks series-ish then? I have encountered it as a Ceasaro Mean in my textbook. If I treat is as a sequence (which i am pretty sure it is, I mean why not) I have managed to show it converges to the same thing as $a_n$. Is this a partial sum of some series? It also reminds me of a Riemann sum, is it related to that?

Did I just show that the arithmetic mean of the sequence $a_n$ converges to the same thing as $a_n$?

I hope these questions make sense.

Thank you

Answer 1

Suppose you have an original sequence $a_1, a_2, \dots$, then we can construct a new sequence $$b_j=\frac{1}{j}\sum_{i=1}^ja_i$$ Just by looking at how $b_j$ is defined, it can clearly be described by the arithmetic average of the first $j$ elements of our sequence $a_i$, i.e $b_j$ is the arithmetic average of $a_1, a_2, \cdots, a_j$. For exampe, if our sequence $a_i$ is defined to be $$a_i=\frac{i-1}{i}$$ then $b_3=\frac13[a_1+a_2+a_3]=\frac13\left[\frac01+\frac12+\frac23\right]=\frac7{18}$.

As with all sequences, we can define the limit of a sequence. For example, if $a_i$ converges, then $\lim_{i \to \infty}a_i$ is defined. Similarly with our new sequence $b_j$ we can define the limit $$\lim_{j \to \infty} b_j=\lim_{j \to \infty}\frac{1}{j}\sum_{i=1}^ja_i$$ So what you have there is just the limit of our new sequence $b_j$.

Answer 2

A series is nothing but the limit of a sequence where the $n$-th term is just the sum of the $n$ first terms of another sequence, like:

$$s_n:= \sum_{k=0}^{n} a_k$$

So the series is defind as:

$$\sum_{k=0}^{\infty} a_k := \lim_{n \to \infty} s_n$$

I assume you are familiar with this construction. Now you only have to replace $s_n$ by the average (rather than the sum) of the first $n$ terms of $(a_n)$ So, no, it's not quite like a series, although both concepts are related

Answer 3

Indeed if $a_n$ converges then $\frac{1}{n}\sum_{i=1}^na_i$ to $a_n$. But the Cesaro mean is used to generalize the notion of convergence. For example consider the sequence

$$a_n=0,1,0,1,0,1,\cdots$$

Does it converge? No. But at some level you might have wanted to say it converges to $0.5$ in the "average" sense. This is what is made strict by the Cesaro mean. You have:

$$\frac{1}{n}\sum_{i=1}^na_i=0.5\text{ if $n$ is even, $0.5(n-1)/n$ if $n$ is odd}$$

And $\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^na_i=0.5$. So now you may actually say $a_n\to0.5$ in the Cesaro sense.

To sum up, the convergence in the Cesaro sense is a generalization of usual convergence. A very important application of this is to talk about the convergence of a Fourier series in the Cesaro sense. But I am not going into any detail here.

Answer 4

The point of this construction is that this new sequence can converge even if the initial sequence does not converge. For instance, take $$ a_k=(-1)^k, $$ which clearly does not converge. The Cesaro mean sequence however converges to $0$.

This construction is used used in proofs concerning the convergence of Fourier series, where the $a_k$ are the partial sums of a Fourier series, $a_k=\sum_{|m|<n} c_me^{imx}$ and the Cesaro mean can be written as $$ \frac1n\sum_{k=1}^n\sum_{-k<m<k}c_me^{imx}=\sum_{-n<m<n}\left(1-\frac{|m|}n\right)c_me^{imx} $$ This dampened coefficient sequence then gives a pointwise limit when inserting the Fourier coefficients $$c_m=\frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-imx}dx.$$ The same insertion into the partial sums gives the Dirichlet kernel which is not exactly helpful for pointwise limit considerations.

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For a nice presentation of the Fourier connection see Carl Offner: "A little harmonic analysis"