Sequences, series, summability Proposition

Cesàro means of a convergent sequence converge

Builds on Cesàro mean Limit of a sequence

Statement

Let $(a_n)_{n \ge 1}$ be a real sequence that converges to a real number $L$ . For $n \ge 1$ , let $\sigma_n = (a_1 + \cdots + a_n)/n$ be its Cesàro mean. Then $\sigma_n \to L$ as $n \to \infty$ .

Remarks

For the proof, let $\varepsilon > 0$ and take $N$ with $|a_k - L| < \varepsilon$ for $k > N$ . For $n > N$ , $|\sigma_n - L| \le \frac{1}{n} \sum_{k=1}^{N} |a_k - L| + \frac{n - N}{n} \varepsilon.$ The first term tends to $0$ , so $|\sigma_n - L| < 2\varepsilon$ for large $n$ . The converse fails. The means of $(-1)^n$ tend to $0$ , but $(-1)^n$ diverges. The statement is a special case of the Stolz–Cesàro theorem with $b_n = n$ .

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Sequence definition
Real sequence definition
Cesàro mean definition
Limit of a sequence definition
Cesàro means of a convergent sequence converge statement

Tag	`0RDRPBMNAC`
URI	`m2://0RDRPBMNAC`
Kind	statement, proposition
Writing	W0 stub
MSC	40G05
Status	This copy of the site has no build data. Run `m2 status 0RDRPBMNAC`.
Uses	Cesàro mean, Limit of a sequence

Cesàro means of a convergent sequence converge

Statement

Remarks

Reading path

Further reading

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