Given a sequence $\{x_n\}$, define its Cesaro Means....` Given a sequence $\{x_n\}_{n=1}^\infty$, define its Cesaro means $y_n= \dfrac{x_1+x_2+\cdots+x_n}{n}$. 
$a)$ If $x_n$ is increasing, show that $y_n$ is increasing. 
This seems obvious to me, so I'm not sure how to do an actual proof that shows this. 
$b)$ If $\lim x_n = L$, show that $\lim y_n= L$ as well. 
I think I need to use the squeeze theorem for this? 
$c)$ Show that the converse of $b)$ may not be true. 
 A: For part $a)$: $$y_{n+1}-y_n = \dfrac{x_1+\cdots + x_n+x_{n+1}}{n+1} - \dfrac{x_1+\cdots + x_n}{n}= \dfrac{nx_{n+1} - (x_1+\cdots + x_n)}{n(n+1)}\geq 0$$
A: To prove a), note
$$y_{n+1} - y_n =  \frac{x_{n+1}}{n+1} - \frac{x_1 + \cdots + x_n}{n(n+1)} \ge \frac{x_{n+1}}{n+1} - \frac{nx_n}{n(n+1)} = \frac{x_{n+1} - x_n}{n+1} \ge 0.$$
To prove b), let $\epsilon > 0$, and choose $n_0 \in \Bbb N$ such that $|x_n - L| < \epsilon$ for all $n\ge n_0$. With this, estimate
$$|y_n - L| \le \frac{1}{n}\sum_{k = 1}^n |x_k - L| < \frac{1}{n} \sum_{k = 1}^{n_0}|x_k - L| + \epsilon \frac{n - n_0}{n}.$$
This implies
$$\limsup_{n\to \infty} |y_n - L| \le \epsilon.$$
Since $\epsilon$ was arbitrary, b) follows.
To prove c), consider $x_n = (-1)^n$. Since 
$$y_{2n} = \frac{x_1 + \cdots + x_{2n}}{2n} = \frac{(-1 + 1) +\cdots + (-1 + 1)}{2n} = 0$$
and 
$$y_{2n+1} = \frac{x_1 + \cdots + x_{2n+1}}{2n+1} = \frac{(-1 + 1) + \cdots + (-1 + 1) - 1}{2n+1} = \frac{-1}{2n+1} \to 0,$$
we deduce $y_n \to 0$. However, $x_n$ is divergent.
A: For part c consider $x_n = (-1)^n$
A: For part $b)$:
$(x_n)$ is convergent and is hence bounded, say $|x_n|\leq M\space \forall n$. Then for $\epsilon >0$, we choose $n$ such that $|x_k-x|<\epsilon$ for $k>n$. Then we fix $n$, so then $y_N=\frac{x_1+\ldots+x_n}{N}+\frac{x_{n+1}+\ldots+x_N}{N}$.
Then we note that for $N>n$ we have that:
$$-\left(\frac{n}{N}\right)\cdot M+\left(\frac{N-n}{N}\right)(x-\epsilon)\leq y_n\leq\left(\frac{n}{N}\right)\cdot M+\left(\frac{N-n}{N}\right)(x+\epsilon)$$
Hence $x-2\epsilon<y_N<x+2\epsilon$.
