Good morning, i have a big problem with this proof. I haven't idea about this proof.


Suppose $\lim_{n\rightarrow\infty}a_n =L $ then $\lim_{n\rightarrow\infty}\ \frac{a_1+a_2+\cdots+a_n}n=L $

I've tried this:

$| a_n-L|<\varepsilon\Rightarrow L-\varepsilon<a_n<L+\varepsilon$ Please help!!


marked as duplicate by Zain Patel, Emily, Mark Viola, colormegone, user223391 Jul 5 '16 at 18:42

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Here's a version of the proof that maybe looks a little longer than other versions, but which has the advantage that each step is more or less obvious:

First, we introduce the notation $$\sigma_n=\frac{a_1+\dots+a_n}{n}.$$

Lemma If there exists $N$ so that $a_n=0$ for all $n>N$ then $\lim\sigma_n=0$.

Proof: $$|\sigma_n|\le\frac{|a_1+\dots+a_N|}{n}.$$QED

Proposition If $\lim a_n=0$ then $\lim\sigma_n=0$.

Proof: Let $\epsilon>0$. Choose $N$ so $|a_n|<\epsilon$ for all $n>N$. Define $$a_n'=\begin{cases}0,&(n\le N), \\a_n,&(n>N).\end{cases}$$The lemma shows that $$\lim(\sigma_n-\sigma_n')=0$$(using what should be obvious notation). But $|a_n'|\le\epsilon$ for every $n$, so $|\sigma_n'|<\epsilon$ for every $n$, hence $$\limsup|\sigma_n'|\le\epsilon.$$Since $\sigma_n-\sigma_n'\to0$ this shows that $\limsup|\sigma_n|\le\epsilon$, and since this holds for every $\epsilon>0$ it follows that $\limsup|\sigma_n|=0$, hence $\lim\sigma_n=0$. QED.

Theorem If $\lim a_n=L$ then $\lim \sigma_n=L$.

Proof: Let $a_n'=a_n-L$. Then the lemma shows that $\lim\sigma_n'=0$, since $\lim a_n'=0$. But $\sigma_n=\sigma_n'+L$, hence $\lim \sigma_n=L$. QED


Let $\epsilon>0$. Therefore, there exists $N \in \mathbb{N}$ such that $m> N \implies a_m < L+\epsilon$. Therefore,

$$\frac{a_1+\cdots+a_n}{n}=\frac{a_1+\cdots +a_N}{n}+\frac{a_{N+1}+\cdots+a_n}{n}$$ $$< \frac{a_1+\cdots +a_N}{n}+\frac{(n-(N+1))(L+\epsilon)}{n}.$$

Passing the $\limsup$ as $n \to \infty$, we get

$$\limsup\limits_{n \to \infty} \frac{a_1+\cdots+a_n}{n} \leq L+\epsilon.$$ Analogously, we can prove $$\liminf\limits_{n \to \infty} \frac{a_1+\cdots+a_n}{n} \geq L-\epsilon.$$ But this holds for every $\epsilon>0.$ Therefore, $$\lim\limits_{n \to \infty} \frac{a_1+\cdots+a_n}{n}=L.$$


It is called Cesaro lemma. First, write in detail your hypothesis. Then, split your sum (what is below) into two parts according to your detailed hypothesis.

Remember that you have to control (ie: make it smaller than $\epsilon$) the quantity : $\mid\frac{1}{n}\sum_{i=1}^n a_i - L \mid = \mid \frac{1}{n}\sum_{i=1}^n (a_i - L )\mid$


We may state Cesàro's lemma in the following form:

If a real sequence $\{a_n\}_{n\geq 0}$ is converging to $L$, it is converging to $L$ on average, too.

The usual $\varepsilon-\delta$ proof is straightforward but quite lengthy. I will give an argument that is easier to follow, at least in my opinion. A converging sequence is a Cauchy sequence, and and averaged Cauchy sequence is still a Cauchy sequence, hence a converging one. Assuming that $$ \lim_{n\to +\infty}\frac{a_1+a_2+\ldots+a_n}{n}= C \neq L $$ we also have $$ \lim_{n\to +\infty}\frac{a_{n+1}+a_{n+2}+\ldots+a_{2n}}{n}=C\neq L $$ or: $$ \lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^{n}\left(a_{n+k}-L\right)\neq 0. $$ That, however, contradicts the fact that $\{a_n\}_{n\geq 0}$ is a Cauchy sequence converging to $L$.

At last, Cesàro's lemma can be seen as a consequence of the Hardy-Littlewood tauberian theorem, too: Karamata's proof is very nice and short.

  • 3
    $\begingroup$ Is this really much shorter, since you have to prove that an averaged sequence of a Cauchy sequence is again Cauchy? $\endgroup$ – Ian Jul 5 '16 at 16:28

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