= Solution
A <sequence> $(x_n)$ is a <convergent sequence> with limit $x$ if
$$
\forall\varepsilon>0\ \exists N\in\mathbb N\
\forall n\ge N:\quad |x_n-x|<\varepsilon.
$$
The index $N$ may depend on $\varepsilon$, but must work for every later term.
Fix $\varepsilon>0$. Choose $N$ such that $|x_i-x|<\varepsilon/2$ for $i\ge N$, and put
$$
C=\sum_{i=1}^{N-1}|x_i-x|.
$$
This is a fixed finite number. For $n\ge N$, the <triangle inequality> gives
$$
\left|\frac1n\sum_{i=1}^nx_i-x\right|
\le\frac1n\sum_{i=1}^n|x_i-x|
\le\frac Cn+\frac{n-N+1}{n}\frac{\varepsilon}{2}
\le\frac Cn+\frac{\varepsilon}{2}.
$$
Choose $n$ sufficiently large that $C/n<\varepsilon/2$ as well. The error is then less than $\varepsilon$, proving
$$
\boxed{\frac1n\sum_{i=1}^nx_i\longrightarrow x.}
$$
\b[Taking successive arithmetic averages preserves the limit.] This is the <Cesàro theorem for convergent sequences>: any troublesome initial terms contribute only a fixed numerator divided by $n$.
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