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?

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

$$\lim \frac{\sum_{k=1}^ny_k}{n}= \lim \frac{\sum_{k=1}^{n+1}y_k-\sum_{k=1}^ny_k}{n+1 - n} = \lim y_{n+1} = y.$$

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 $|S_n - y| < \epsilon$.