Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Two Limits Equal - Proof
Prove convergence of the sequence $(z_1+z_2+\cdots + z_n)/n$ of Cesaro means

Suppose $a_n \rightarrow c$ as $n \rightarrow \infty$. We want to show that $\dfrac{1}{n} \sum_{i=1}^n a_n \rightarrow c$ as $n \rightarrow \infty$.

I know I can use the definition to prove this by showing that for any given $\epsilon$, we can find an $N$ such that whenever $n \ge N$, we can have $|\dfrac{1}{n} \sum_{i=1}^n a_n-c| \le \epsilon$.

I am wondering whether there is another easier way to prove this.

Thank you very much. Hanna

share|improve this question

marked as duplicate by Paul, Martin Sleziak, Martin Argerami, Austin Mohr, Mhenni Benghorbal Dec 4 '12 at 7:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
This should help. –  Pedro Tamaroff Dec 4 '12 at 5:05

2 Answers 2

Let $\epsilon > 0$. Select $N$ so that $n \geq N$ implies $|a_n - c| < \epsilon$. The idea is that we would like to break up the average $\frac{1}{n}\sum_{i=1}^n a_i$ into two parts. The estimate is this

\begin{align*} \bigg|\frac{1}{n}\sum_{i=1}^n a_i - c \bigg| &\leq \frac{\sum_{i=1}^N |a_i - c|}{n} + \frac{\sum_{i=N+1}^n |a_i - c|}{n} \\ &\leq \frac{\sum_{i=1}^N |a_i - c|}{n} + \frac{\sum_{i=N+1}^n \epsilon}{n} \\ &= \frac{\sum_{i=1}^N |a_i - c|}{n} + \frac{(n- (N+1))\epsilon}{n} \end{align*}

Now, the left term has a constant numerator, and so goes to zero. The right term approaches $\epsilon$ as $n \rightarrow \infty$. Note that $N$ is held fixed as $n$ goes to $\infty$. If you like, you can be more rigorous by choosing $M$ so that $n \geq M$ implies, say, that the left term is less than $\epsilon$ and the right quotient $\frac{n - (N+1)}{n}$ is less than $1 + \epsilon$. This gives that for $n\geq \max\{M,N\}$, the entire thing is bounded by $\epsilon + (1 + \epsilon)\epsilon = \epsilon^2 + 2\epsilon$, which can be made arbitrarily small.

share|improve this answer

I personally think using the definition works best here.

Say $\lim \;a_n =\ell $. Then

$$\left|\frac 1 n \sum_{k=1}^n a_k-\ell \right|=$$

$$\left|\frac 1 n \sum_{k=1}^n a_k-\frac 1 n \sum_{k=1}^n \ell \right|=$$

$$\frac 1 n\left| \sum_{k=1}^n (a_k- \ell) \right|$$

Now, we know that:

$(a)$ we can make $|a_k-\ell| $ small by taking $k$ sufficiently large

$(b)$ we can make, for any $M$, $M/n$ small by taking $n$ large.

The idea now is to split the sum in two: a tail, that we will make small by using $(a)$, and some first terms, which we will make small by using $(b)$. Let $\epsilon >0$ be given. Let $N$ be such that $|a_k-\ell|<\epsilon/2$ when $n\geq N$. So

$$\frac 1 n\left| \sum_{k=1}^n (a_k- \ell) \right|\leq \frac 1 n \sum_{k=1}^{N} |a_k- \ell| +\frac 1 n \sum_{k=N+1}^n |a_k- \ell| $$

Set $\Phi_N=\max\limits_{1\leq k \leq N}|a_k-\ell|$. Note that $\Phi_N$ gets fixed for every choice of $N$ we make, and it can't get too large. In fact, since $a_k\to \ell$, we must have that $\Phi_N$ is not larger than $\sup |a_k-\ell|=\Phi$.

Since

$$\frac 1 n \sum_{k=N+1}^n |a_k- \ell|<\frac 1 n\frac \epsilon 2 \sum_{k=N+1}^n 1=\\=\frac {n-N} n\frac \epsilon 2$$

then

$$\frac 1 n\left| \sum_{k=1}^n (a_k- \ell) \right|<\frac N n \Phi -\frac \epsilon 2 \frac {N}{n}+\frac \epsilon 2 $$

Now it is all done take $N_1$ so that for $n>N_1$ $$\eqalign{ & \frac{{N\Phi}}{n} < {\epsilon}/2 \cr & \frac{N}{n} < {1} \cr} $$

Then, for $n>N_2=\max(N,N_1)$ we will certainly have

$$\frac 1 n\left| \sum_{k=1}^n (a_k- \ell) \right|<\frac N n \Phi -\frac \epsilon 2 \frac {N}{n}+\frac \epsilon 2 <\frac \epsilon 2+\frac \epsilon 2=\epsilon $$

so that $\hat a_n \to \ell$. This nice exercise is telling you that every summable sequence is also Cèsaro summable, by the way.

share|improve this answer
    
Thanks very much for the reply. I have a question about the last inequality on the last expression above. We choose N1 such that $N/n < 1$ if $n \ge N_1$. So $-\dfrac{\epsilon}{2}\dfrac{N}{n} > -\dfrac{\epsilon}{2}$. So it seems that the last inequality is wrong. –  Hanna Dec 6 '12 at 3:42
    
@Hanna You're right. Then you can just go with $-\frac \epsilon 2 \frac N n<0$ –  Pedro Tamaroff Dec 6 '12 at 15:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.