Two Limits Equal - Proof that $\lim_{n\to\infty }a_n=L$ implies $\lim_{n\to\infty }\frac{\sum_1^na_k}n=L$ [duplicate]

Here is the correct proof: Given any $\epsilon>0$, since $\lim_{n\to \infty }a_n=L$, there exists $N_0\in\mathbb{N}$ such that $$|a_k-L|<\frac{\epsilon}{2}\mbox{ for }k\geq N_0.$$ Now, for the given $\epsilon$ and $N_0$, we can choose an integer $N_1$ large enough such that $$\sum_{k=1}^{N_0}|a_k|+N_0|L|<\frac{N_1\epsilon}{2}.$$ Hence, for $n\geq N_2:=\max\{N_0,N_1\}$, we have $$|m_n-L|=\Big|\frac{\sum_{k=1}^na_k}{n}-L\Big|=\Big|\frac{\sum_{k=1}^n(a_k-L)}{n}\Big| \leq\frac{\sum_{k=1}^{N_0}|a_k-L|}{n}+\frac{\sum_{k=N_0+1}^n|a_k-L|}{n}:=I+II<\epsilon,$$ because $$I=\frac{\sum_{k=1}^{N_0}|a_k-L|}{n}\leq\frac{\sum_{k=1}^{N_0}|a_k|+N_0|L|}{N_2}\leq\frac{\sum_{k=1}^{N_0}|a_k|+N_0|L|}{N_1}<\frac{\epsilon}{2}$$ and $$II=\frac{\sum_{k=N_0+1}^n|a_k-L|}{n}<\frac{\sum_{k=N_0+1}^n\epsilon/2}{n}=\frac{(n-N_0)\epsilon/2}{n}\leq\frac{\epsilon}{2}.$$ Therefore, by definition, we have $$\lim_{n\to \infty } m_n=L$$