[proofplan] We split the Ces\`aro mean $\sigma_N - L$ into a "head" (finitely many early terms) and a "tail" (terms close to $L$). The head is a fixed finite sum divided by $N+1$, which tends to zero. The tail has $N - M + 1$ terms each within $\varepsilon$ of $L$, so its average is within $\varepsilon$. The triangle inequality combines the two bounds. [/proofplan] [step:Split $\sigma_N - L$ into head and tail sums] Let $\varepsilon > 0$. Since $s_n \to L$, there exists $M \in \mathbb{N}$ such that $|s_n - L| < \varepsilon$ for all $n \geq M$. Write: \begin{align*} \sigma_N - L = \frac{1}{N+1}\sum_{n=0}^N (s_n - L) = \underbrace{\frac{1}{N+1}\sum_{n=0}^{M-1}(s_n - L)}_{\text{head}} + \underbrace{\frac{1}{N+1}\sum_{n=M}^N(s_n - L)}_{\text{tail}}. \end{align*} [/step] [step:Bound the head term by dilution] The sum $A := \left|\sum_{n=0}^{M-1}(s_n - L)\right|$ is a fixed finite quantity independent of $N$. Therefore: \begin{align*} \left|\frac{1}{N+1}\sum_{n=0}^{M-1}(s_n - L)\right| \leq \frac{A}{N+1}. \end{align*} Choose $N_0 \geq M$ such that $A/(N_0 + 1) < \varepsilon$. For all $N \geq N_0$, the head is less than $\varepsilon$. [/step] [step:Bound the tail term using the convergence hypothesis] For $n \geq M$, $|s_n - L| < \varepsilon$. The tail has $N - M + 1$ terms: \begin{align*} \left|\frac{1}{N+1}\sum_{n=M}^N(s_n - L)\right| \leq \frac{1}{N+1}\sum_{n=M}^N |s_n - L| < \frac{(N - M + 1)\varepsilon}{N+1} \leq \varepsilon. \end{align*} [/step] [step:Combine by the triangle inequality to conclude $\sigma_N \to L$] For all $N \geq N_0$: \begin{align*} |\sigma_N - L| \leq \frac{A}{N+1} + \frac{(N-M+1)\varepsilon}{N+1} < \varepsilon + \varepsilon = 2\varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\sigma_N \to L$. [/step]