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Let $(a_n)$ be a sequence of real numbers and $g\in \mathbb{R} $. Assume that $$ \frac{a_1+\cdots+a_n}{n}\rightarrow g$$ as $n \rightarrow \infty$. Now let $k \in \mathbb{N}$. Is it then $$\frac{a_{k+1}+\cdots+a_{k+n}}{n} \rightarrow g $$

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Yes: $$\frac{a_{k+1}+\cdots+a_{k+n}}n=\frac{k+n}n\left(\frac{a_1+\cdots+a_{k+n}}{k+n}-\frac{a_1+\cdots+a_k}{k+n}\right)\;,$$ which clearly approaches $g$ as $n\to\infty$.

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Yes this is true since $$\lim_{n\rightarrow\infty}\frac{a_1+a_2+\cdots +a_k}{n}=0,$$ as the numerator is a fixed constant since $k$ is fixed. This means we can remove the first $k$ terms without affecting the limit.

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May I ask you what happened to $a_{k+n}$ in your limit above? –  B. S. Jan 18 '12 at 19:04

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