= Solution
The <series> $\sum_{n\ge1}a_n$ converges to $A$ when its <partial sums> $s_n$ have <limit of a sequence> $A$: for every $\varepsilon>0$, there is $N$ such that $|s_n-A|<\varepsilon$ for all $n\ge N$.
Its averages are the <Cesaro means> of the sequence of <partial sums>. To prove their convergence, choose $N$ with $|s_k-A|<\varepsilon/2$ for $k\ge N$. Then
$$
\left|\frac1n\sum_{k=1}^n s_k-A\right|
\le\frac1n\sum_{k=1}^{N-1}|s_k-A|+\frac{n-N+1}{n}\frac\varepsilon2.
$$
The finite first sum divided by $n$ tends to zero. This proves the <Cesaro theorem for convergent sequences> in this setting, and
$$
\boxed{\widetilde A=A.}
$$
The converse fails. Take $a_n=(-1)^{n+1}$, so $s_n=1$ for <odd> $n$ and $s_n=0$ for <even> $n$. The <partial sums> do not converge, but
$$
\boxed{\frac{s_1+\cdots+s_n}{n}=\frac{\lceil n/2\rceil}{n}\longrightarrow\frac12.}
$$
Thus the alternating <series> is <Cesàro summable> without having an ordinary sum.
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