How to compute the above limit Let $y_n$ be a sequence such that  $\lim_{n\to \infty} y_n= y$ then prove that

$ \lim \dfrac{y_1+y_2+...+y_n}{n}= y$

My try:
As  $\lim_{n\to \infty} y_n= y$ then given $\epsilon>0$ there exists $N$ such that $\forall n\ge N ;||y_n-y||<\epsilon $.
Now $||\dfrac{y_1+y_2+...+y_n}{n}-y||\le n^{-1}\{||y_1-y||+||y_2-y||+...+||y_n-y||\}$
How to make this limit tend to zero?
 A: Note that
$$S_n = \frac1{n} \sum_{k=1}^n y_k = \frac1{n} \sum_{k=1}^N y_k+  \frac1{n} \sum_{k=N+1}^n y_k$$
Now use that there exists $N$ such that $y - \epsilon < y_k < y + \epsilon$ for $k > N.$
Hence
$$\frac1{n}\sum_{k=1}^N y_k+(y- \epsilon)(1 - N/n)< S_n < \frac1{n}\sum_{k=1}^N y_k+(y+ \epsilon)(1 - N/n) $$
With $N$ fixed take $\limsup$ and $\liminf$ as $n \to \infty$ to find for all $\epsilon > 0$
$$y - \epsilon \leqslant \liminf S_n \leqslant \limsup S_n < y + \epsilon.$$
Since, $\epsilon$ can be arbitrarily small we have
$$\liminf S_n = \limsup S_n =\lim S_n = y.$$
Alternatively, this is just a special case of the Stolz-Cesaro theorem
$$\lim \frac{\sum_{k=1}^ny_k}{n}= \lim \frac{\sum_{k=1}^{n+1}y_k-\sum_{k=1}^ny_k}{n+1 - n} = \lim y_{n+1} = y.$$
Addendum
You might not like the appearance of $\liminf$ and $\limsup$ here. You can proceed with your argument. Choose fixed $N$ such that $|y_k - y| < \epsilon/2$ for $k > N$, and we have
$$|S_n - y|  = \left|\frac1{n} \sum_{k=1}^N (y_k-y)+  \frac1{n} \sum_{k=N+1}^n (y_k-y)\right| \\ \leqslant \frac1{n} \sum_{k=1}^N |y_k-y|+  \frac1{n} \sum_{k=N+1}^n |y_k-y| \\ \leqslant \frac1{n} \sum_{k=1}^N |y_k-y| +  (1 - N/n)\epsilon/2 \\ \leqslant \frac1{n} \sum_{k=1}^N |y_k-y| + \epsilon/2.$$
There exists $N'$ such that if $n > \max(N, N')$ we have the first term on the RHS less than $\epsilon/2$ and $|S_n - y| < \epsilon$.
