Assume that $\lim x_n = L$ and define $(y_n)$ by $$y_n = \frac{1}{n} \sum_{k=1}^{n}x_k.$$
(a) Prove that $\lim y_n = L$
(b) Is it possible that $(y_n)$ converges, even if $(x_n)$ does not?
Any ideas on how to get started on part (a)?
In the afternoon section we briefly discussed how to get this started. I haven't looked at it too in depth yet but I'll share the few notes I have.
Suppose $a_n \rightarrow L$
$$\left|\frac{a_1 + a_2 + a_3 + \ldots + a_N + a_{N+1} + \ldots + a_n}{n} - L\right|$$
$$\leq \frac{\left| a_1+a_2+ \ldots +a_N \right|}{n} + \left|\frac{a_{N+1}+a_{N+2}+ \ldots + a_n}{n-N} \cdot \left(\frac{n-N}{n}\right) - L\right|$$
I also have a mysterious $L-\varepsilon$, $L+\varepsilon$ floating around underneath as well but I imagine that's because at some point you get that you need to subtract/add epsilon from what the sequence converges to near your final steps.
If anyone else can chime in with anything, that would be much appreciated! 
I'm still not sure how to go about solving this, but I found some resources that gives more insight into the problem and its importance.
Has anyone figured this out yet?
One Idea that I have is this:
we know for sure that the quantities $n - N, \frac{n - N}{n} \geq 0$ and that $|a - |L|| \leq |a - L|$ .
It seems that together this means that $\frac{\left|a_{1} + ... + a_{N}\right|}{n} + \left|\frac{a_{N + 1} + ... + a_{n}}{n - N}\cdot\frac{n - N}{n} - L\right| \leq \frac{\left|a_{1} + ... + a_{N}\right|}{n} + \frac{\left|a_{N + 1} + ... + a_{n}\right|}{n - N}\cdot\frac{n - N}{n} - \left|L\right|$ Not sure if that's what we need or not though...