Mean of a Convergent Sequence 
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
 A: 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.
A: 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.
