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 $\lim_{n\to \infty } m_n=L$