Cesàro means of a convergent sequence converge

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 $b_n = n$.

Explanations

L1 human

If the terms settle near $L$, then the running average also settles near $L$. The early terms that are far from $L$ count less and less, because we divide by $n$.