# Math Help - sequence, convergence

1. ## sequence, convergence

Suppose that $(x_n)$ is a sequence in $\mathbb{R}$. Define a sequence $(y_n)$ by $y_n=\frac{x_1+x_2+\cdots+x_n}{n}$, $\forall n \in \mathbb{N}$. Prove that if $(x_n)$ converges to $x \in \mathbb{R}$ then $(y_n)$ converges to $x$.

2. Hello,
Originally Posted by xboxlive89128
Suppose that $(x_n)$ is a sequence in $\mathbb{R}$. Define a sequence $(y_n)$ by $y_n=\frac{x_1+x_2+\cdots+x_n}{n}$, $\forall n \in \mathbb{N}$. Prove that if $(x_n)$ converges to $x \in \mathbb{R}$ then $(y_n)$ converges to $x$.
This is Cesaro's mean. And you want to prove Cesaro's lemma.
There is a proof in the French Wikipedia (Lemme de Cesàro - Wikipédia), but not in the English one.
I'll translate and adapt it here...

$x_n \to x$ can be translated this way :
$\forall \varepsilon >0,~ \exists N \in \mathbb{N},~ \forall n>N,~ |x_n-x|< \varepsilon/2$

Let $S_N=\sum_{k=1}^N |x_k-x|$

Let's assume n>N :
$|y_n-x|=\left|\frac 1n \sum_{k=1}^n (x_k-x)\right|$
By the triangle inequality, we have :
$|y_n-x| \leqslant \frac 1n \sum_{k=1}^N |x_k-x|+\frac 1n \sum_{k=N+1}^n |x_k-x|$
$|y_n-x| \leqslant \frac{S_N}{n}+\frac{n-N}{n} (\varepsilon/2)=\frac{S_N}{n}+\varepsilon/2 -\frac{N}{n} (\varepsilon/2) \leqslant \frac{S_N}{n}+\varepsilon/2$
(because N/n (epsilon/2) >0)

But $S_N$ doesn't depend on n. So $\lim_{n \to \infty} \frac{S_N}{n}=0$
This means that for any $\varepsilon >0$, there exists an integer N' such that for all n>N', we have :
$\left|\frac{S_N}{n}\right|\leqslant \varepsilon/2$

By combining this latter inequality to the previous one, we have, for $n>\max\{N,N'\}$ :
$|y_n-x| \leqslant \varepsilon/2+\varepsilon/2=\varepsilon \qquad \square$