Solution
= Solution
Let $d_j=a_{j+1}-a_j$ and suppose $d_j\to L$. Telescoping gives
$$
\frac{a_n}{n}=\frac{a_1}{n}+\frac1n\sum_{j=1}^{n-1}d_j.
$$
To justify the <Cesaro mean> limit directly, choose $J$ so that $|d_j-L|<\varepsilon$ for $j\geq J$. The finite initial sum contributes at most $n^{-1}\sum_{j<J}|d_j-L|$, which tends to zero. The remaining terms contribute at most $\varepsilon$, and replacing $(n-1)L/n$ by $L$ contributes $|L|/n$. Consequently \b[$\boxed{a_n/n\to L}$].