Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $L$. – Arturo Magidin Nov 20 '11 at 5:36
@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... – Arturo Magidin Nov 20 '11 at 5:37
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? – Arturo Magidin Nov 20 '11 at 5:42
Another thing which might be worth mentioning is that the sequence $(m_n)$ is called Cesàro mean of the sequence $(a_n)$. – Martin Sleziak Nov 20 '11 at 10:38
up vote 5 down vote accepted

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.