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If $\{s_n\}$ is a complex sequence, define its arithmetic means $\sigma_n$ by $$\sigma_n=\dfrac{s_0+s_1+\dots+s_n}{n+1}.$$ If $\lim s_n=s$, prove that $\lim \sigma_n=s.$

My proof: Let $t_n:=s_n-s$ and $\tilde{\sigma}_n:=\sigma_n-s$. Then $\lim t_n=0.$ We must prove that $\lim \tilde{\sigma}_n=0.$

For any $\varepsilon>0$, there exists $N_{\varepsilon}$ such that $n\geqslant N_{\varepsilon}$ implies $|t_n|<\varepsilon$. We have $$|\tilde{\sigma}_n|=\dfrac{|t_0+t_1+\dots+t_n|}{n+1}\leqslant \dfrac{|t_0+\dots+t_N|+|t_{N+1}|+\dots+|t_n|}{n+1}<\dfrac{|t_0+\dots+t_N|+\varepsilon(n-N)}{n};$$ Keeping $N$ fixed, and letting $n\to\infty$, we get $$\limsup_{n\to \infty}|\tilde{\sigma}_n|\leqslant \varepsilon.$$ Since $\varepsilon$ is arbitrary, $\limsup_{n\to\infty}|\tilde{\sigma}_n|=0$. It is easy to check that $\lim|\tilde{\sigma}_n|=\lim\tilde{\sigma}_n=0$. Hence $\tilde{\sigma}_n$ converges to zero. Q.E.D

Can anyone check my proof?

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    $\begingroup$ sounds good, well done! $\endgroup$ – Math-fun Aug 25 '15 at 18:28
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    $\begingroup$ Right. This is a standard method of dividing a sum into a "bad" part (usually the initial part, which, since it is fixed, can be ignored when it is divided by $n$), and a "good" part where things behave well. $\endgroup$ – marty cohen Aug 25 '15 at 19:20
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Yes, apart from a few typos, your proof is correct.

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