# Arithmetic mean sequence of a convergent complex sequence

Let $\{s_n\}$ be a complex sequence. Let $\sigma_n = (\sum_{i=0}^n s_i)$/$(n+1)$.

Say $s_n → s$.

I have tried and i could only show that $|\sigma_n| →|s|$.

How do i show that $\sigma_n →s$?

I have searched for it and some solutions use "inequality of complex number", but as you know order relation cannot be defined in $\mathbb{C}$.

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@Brian It's a typo. Dividing by n+1 is what i want! – Katlus Oct 5 '12 at 11:43

Hint:

1. Fix $\varepsilon>0$ and choose $N=N_\varepsilon$ such that $$|s_n-s|<\varepsilon$$ for $n>N$, and note that $$|\sigma_n - s|=\left|\frac{1}{n+1}\sum_0^n (s_k-s)\right|\leq\frac{1}{n+1}\sum_0^n |s_k-s|\tag{1}$$
2. Break up the sum in (1) at $k=N$.

3. Use the max esitmate on the first terms and the $\varepsilon$-estimate on the tail.

Do you see how to complete the proof?

A sequence $s_n$ for which $\sigma_n$ converges is called Cesàro-summable, the result is that convergence implies Cesàro-summable.

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I'm not really good in LaTeX. There's "Math Processing Error" after "Note that" on my phone.. It's awkward but would you please simplify that? – Katlus Oct 5 '12 at 12:15
$|\sigma_n - s|=\left|\frac{1}{n+1}\sum_0^n (s_k-s)\right|\leq\frac{1}{n+1}\sum_0^n |s_k-s|\tag{1}$ – Katlus Oct 5 '12 at 12:18
Maybe the tag "(1)" is the problem? Can you read this..? $$|\sigma_n - s|=\left|\frac{1}{n+1}\sum_0^n (s_k-s)\right|\leq\frac{1}{n+1}\sum_0^n |s_k-s|$$ – AD. Oct 5 '12 at 12:34
Yes, i got it thank you :) – Katlus Oct 5 '12 at 14:18