If $\{s_n\}$ is a sequence of positive real numbers such that $\lim_{n \rightarrow \infty }s_n = s$, is it true that $\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^n s_k =s $ as well?
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edited it to read "a sequence of positive real numbers" instead of just "a sequence of real numbers" – Rasputin Oct 18 '18 at 07:48
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This is indeed true. – General Grievous Oct 18 '18 at 07:50
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This is true. Let $\varepsilon>0$ be given, choose $N>0$ such that $n\geq N$ implies that $|s_n-s|<\varepsilon$ then for $M>N$
$$\left|\sum_{k=1}^{M}\frac{s_k}{M}-s\right|=\left|\sum_{k=1}^{M}\frac{s_k-M\cdot s}{M}\right|=\left|\frac{1}{M}\sum_{k=1}^{M}(s_k-s)\right|\leq \frac{1}{M}\left|\sum_{k=1}^{N}(s_k-s)\right|+\frac{1}{M}\sum_{k=N+1}^{M}\varepsilon$$
Do you see how one could proceed from this?
OgvRubin
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1Yeah, letting $M \rightarrow \infty$ makes the right-most side approach something $\le \varepsilon$ – Rasputin Oct 18 '18 at 07:55