= Solution
Suppose $a_n\to L$. Given $\varepsilon>0$, choose $N$ such that $|a_k-L|<\varepsilon/2$ for $k>N$. Then
$$
\left|\frac1n\sum_{k=1}^na_k-L\right|
\leq
\frac1n\sum_{k=1}^N|a_k-L|
+\frac1n\sum_{k=N+1}^n|a_k-L|.
$$
The first term tends to zero because its numerator is fixed, and the second is at most $\varepsilon/2$. Thus the <Cesaro mean> tends to $L$.
The converse is false. For $a_n=(-1)^n$, the Cesaro means tend to $0$, while the original <sequence> alternates between $-1$ and $1$ and does not converge.
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