Cesàro Mean

From ProofWiki
Jump to: navigation, search

Theorem

Let $\left\langle{a_n}\right\rangle$ be a sequence of complex numbers.

Suppose that $\left\langle{a_n}\right\rangle$ converges to $\ell$ in $\C$:

$\displaystyle \lim_{n \to \infty} a_n = \ell$


Then also:

$\displaystyle \lim_{n \to \infty} \frac{a_1 + \dots + a_n} n = \ell$


Proof

For every fixed integer $n_0$, we write:

$\displaystyle \left\vert{\frac{a_1 + \dots + a_n} n - \ell}\right\vert \le \frac{\left\vert{a_1 - \ell}\right\vert + \dots + \left\vert{a_n - \ell}\right\vert} {n} \le \frac {n_0 \sup_{k \le n_0} \left\vert{a_k - \ell}\right\vert} n + \sup_{n_0 < k \le n} \left\vert{a_k - \ell}\right\vert$

As $n$ tends to $+\infty$, we get:

$\displaystyle \limsup_{n \to \infty} \left\vert{\frac{a_1 + \dots + a_n} n - \ell}\right\vert \le \sup_{k > n_0} \left\vert{a_k - \ell}\right\vert$

As $n_0$ tends to $+\infty$, we finally conclude:

$\displaystyle \limsup_{n \to \infty} \left\vert{\frac{a_1 + \dots + a_n} n - \ell}\right\vert = 0$

$\blacksquare$


Remarks

  • When working with an arbitrary sequence $\left\langle{a_n}\right\rangle$ of real numbers, the same truncation trick leads to:
$\displaystyle \liminf_{n \to \infty} a_n \le \liminf_{n\to\infty} \frac{a_1 + \dots + a_n} n \le \limsup_{n \to \infty} \frac{a_1 + \dots + a_n} n \le \limsup_{n \to \infty} a_n$

As a corollary, the conclusion of the theorem holds in the real case when $\ell = \pm \infty$.


Source of Name

This entry was named for Ernesto Cesàro.