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Problem: Given that $\lim_{n\to\infty }a_n=L$ and $m_n=\frac{\sum_{1}^{n}a_k}{n}$. Prove that $\lim m_n=L$

Proof: We have $\sum_{1}^{n}a_k=na_k$, so $m_n=\frac{\sum_{1}^{n}a_k}{n}=\frac{na_k}{n}=a_k$ and $\lim a_k=L.\square$

Is this a correct proof? I am confused with the subscripts $m$ and $k$ and not sure if I am using the right one in the right spot, in the proof. Thanks.

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    This question is not a duplicate, since here the OP is asking if a particular proof is valid. In the "duplicate" question, the OP is asking for any proof. (Though the fact that the OP here accepted an alternate proof muddies the issue.)2016-01-14

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