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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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    $\begingroup$ I think that the index on the summation is supposed to be $k$. That is, "$\sum_1^n a_k$" means $a_1+a_2+\cdots+a_n$, not $a_k+a_k+\cdots+a_k$ ($n$ summands); so you don't have that the sum is equal to $na_k$. Otherwise, what you write is false, because if $k$ is fixed, then $\lim_{n\to\infty}a_k = a_k$, not $L$. $\endgroup$ – Arturo Magidin Nov 20 '11 at 5:36
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    $\begingroup$ @Bill Cook: Since the confusion arises from the lack of indices in what he wrote, I would think it best for the OP to understand their importance rather than for you to add them "in spite" of him... $\endgroup$ – Arturo Magidin Nov 20 '11 at 5:37
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    $\begingroup$ Yes, the lack of indices/subscripts makes the issue confusing. That's why they are important! It's important to know what the indices mean; for example, the summation has a "hidden index" which you did not write. It should "really" be $$\sum_{k=1}^n a_k.$$ Was it written this way in your assignment? $\endgroup$ – Arturo Magidin Nov 20 '11 at 5:42
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    $\begingroup$ Another thing which might be worth mentioning is that the sequence $(m_n)$ is called Cesàro mean of the sequence $(a_n)$. $\endgroup$ – Martin Sleziak Nov 20 '11 at 10:38
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    $\begingroup$ 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.) $\endgroup$ – Rory Daulton Jan 14 '16 at 23:40
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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 $$\lim_{n\to \infty } m_n=L$$

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