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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?

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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.

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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

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