Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $x_n \to L$. Prove that $$\lim_{N\to\infty}\dfrac1N\sum_{n=1}^Nx_n=L.$$

My idea is, write $L=\sum_{n=0}^{N}\frac{L}{N}$. So we have, $\lim_{\ n\to\infty}\frac{1}{N}\sum_{n=0}^{N}(x_n-L)=\lim_{\ n\to\infty}\frac{1}{N}\sum_{n=0}^{N}(x_n-L)+\frac{1}{N}\sum_{m=N_0+1}^{N}(x_m-L)$. Here the LH sum tends to $0$ for $N$ large enough. But I don't know where to go from here?

share|improve this question

1 Answer 1

Hint: Let $\varepsilon>0$ be given arbitarily. As $x_n\to L$, there must exists some $N\in\mathbb{N}$ such that $|x_n-L|<\varepsilon/3$ for all $n\ge N$. Now we consider the following formula: $$\bigg|\frac{x_1+\cdots+x_N+x_{N+1}+\cdots+x_{N+M}}{N+M}-L\bigg|,$$ where $M$ is a strictly positive integer. Notice that \begin{align*} \bigg|\frac{x_1+\cdots+x_N+x_{N+1}+\cdots+x_{N+M}}{N+M}-L\bigg|&=\bigg|\frac{\sum_{i=1}^N x_i}{N+M}+\frac{\sum_{j=1}^M(x_{N+j}-L)-NL}{N+M}\bigg|\\ &\le \frac{1}{N+M}\cdot \Big|\sum_{i=1}^N x_i\Big|+\frac{1}{N+M}\cdot \bigg(\sum_{j=1}^M|x_{N+j}-L|+N\,|L|\bigg)\\ &<\frac{1}{N+M}\cdot \Big|\sum_{i=1}^N x_i\Big|+\frac{1}{N+M}\bigg(M\cdot\frac{\varepsilon}{3}+N\,|L|\bigg)\\ &=\frac{1}{N+M}\cdot \Big|\sum_{i=1}^N x_i\Big|+\frac{\varepsilon/3}{\frac{N}{M}+1}+\frac{|L|}{1+\frac{M}{N}}\\ &<\frac{1}{N+M}\cdot \Big|\sum_{i=1}^N x_i\Big|+\frac{\varepsilon}{3}+\frac{|L|}{1+\frac{M}{N}}. \end{align*} Now you can see that as $M$ large enough, the term $\frac{|L|}{1+\frac{M}{N}}$ and the term $\frac{1}{N+M}\cdot \Big|\sum_{i=1}^N x_i\Big|$ can always be made smaller than $\varepsilon/3$ (note that $N$ is fixed). Conclude.

share|improve this answer
I'm sorry, I'm not following. –  user96137 Dec 11 '13 at 8:32
Where did you get stuck? –  user65018 Dec 16 '13 at 22:09

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.