Cesàro Mean

Theorem
Let $\left\langle {a_n}\right\rangle$ be a sequence of complex numbers. If $\left\langle {a_n} \right\rangle$ converges to $\ell$ in $\mathbb{C}$, then $\displaystyle\frac{a_1 + \dots + a_n}{n} \xrightarrow[n \to \infty]{} \ell$.

Proof
For every fixed integer $n_0$, we write:
 * $\displaystyle\left|\frac{a_1 + \dots + a_n}{n} - \ell\right| \leq \frac{\left|{a_1 - \ell}\right| + \dots + \left|{a_n - \ell}\right|} {n} \leq \frac{\sup_{k \leq n_0} \left|{a_k-\ell}\right|}{n} + \sup_{n_0 < k \leq n} \left|{a_k - \ell}\right|$

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

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

As $n_0$ tends to $+\infty$, we finally conclude:
 * $\displaystyle\limsup_{n\to\infty} \left|{\frac{a_1 + \dots + a_n}{n} - \ell}\right| = 0$

Remarks

 * The theorem and its proof hold in any normed vector space.


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

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