[proofplan] We prove that the arithmetic means of the partial sums converge to the same limit as the partial sums themselves. The argument separates the Cesaro average into a finite initial block and a tail block. The initial block becomes negligible because it is divided by $N+1$, while the tail block is uniformly close to $s$ because the partial sums converge to $s$. [/proofplan] [step:Reduce the Cesaro mean to an average of errors] For each $n \in \mathbb{N} \cup \{0\}$, define the error term $e_n \in \mathbb{R}$ by \begin{align*} e_n := s_n - s. \end{align*} Since $s_n \to s$ as $n \to \infty$, we have $e_n \to 0$ as $n \to \infty$. For each $N \in \mathbb{N} \cup \{0\}$, \begin{align*} \sigma_N - s &= \frac{1}{N+1}\sum_{n=0}^{N} s_n - s \\ &= \frac{1}{N+1}\sum_{n=0}^{N} s_n - \frac{1}{N+1}\sum_{n=0}^{N} s \\ &= \frac{1}{N+1}\sum_{n=0}^{N} (s_n - s) \\ &= \frac{1}{N+1}\sum_{n=0}^{N} e_n. \end{align*} Thus it suffices to prove that \begin{align*} \frac{1}{N+1}\sum_{n=0}^{N} e_n \to 0 \end{align*} as $N \to \infty$. [/step] [step:Choose a tail where all errors are small] Let $\varepsilon > 0$. Since $e_n \to 0$, there exists an integer $M \geq 0$ such that \begin{align*} |e_n| < \frac{\varepsilon}{2} \end{align*} for every integer $n \geq M$. Define the finite initial-error constant $A \in [0,\infty)$ by \begin{align*} A := \sum_{n=0}^{M-1} |e_n|, \end{align*} with the convention that $A = 0$ if $M = 0$. [/step] [step:Bound the initial block and the tail block separately] For every integer $N \geq M$, the triangle inequality gives \begin{align*} \left|\frac{1}{N+1}\sum_{n=0}^{N} e_n\right| &\leq \frac{1}{N+1}\sum_{n=0}^{N} |e_n| \\ &= \frac{1}{N+1}\sum_{n=0}^{M-1} |e_n| + \frac{1}{N+1}\sum_{n=M}^{N} |e_n| \\ &\leq \frac{A}{N+1} + \frac{1}{N+1}\sum_{n=M}^{N} \frac{\varepsilon}{2} \\ &= \frac{A}{N+1} + \frac{N-M+1}{N+1}\cdot \frac{\varepsilon}{2} \\ &\leq \frac{A}{N+1} + \frac{\varepsilon}{2}. \end{align*} Choose an integer $N_0 \geq M$ such that \begin{align*} \frac{A}{N_0+1} < \frac{\varepsilon}{2}. \end{align*} Then for every integer $N \geq N_0$, since $N+1 \geq N_0+1$, we have \begin{align*} \left|\frac{1}{N+1}\sum_{n=0}^{N} e_n\right| < \varepsilon. \end{align*} [guided] We want to prove that the average of the errors tends to $0$. The convergence $e_n \to 0$ tells us that all sufficiently late errors are small, but it gives no smallness for the finitely many early errors. The key point is that finitely many fixed terms become harmless after division by $N+1$. Let $\varepsilon > 0$. Since $e_n \to 0$, there exists an integer $M \geq 0$ such that \begin{align*} |e_n| < \frac{\varepsilon}{2} \end{align*} for every integer $n \geq M$. The early errors are the finitely many numbers $e_0,\dots,e_{M-1}$. Define \begin{align*} A := \sum_{n=0}^{M-1} |e_n|, \end{align*} with $A = 0$ when $M = 0$. This is a finite real number because it is a finite sum of [real numbers](/page/Real%20Numbers). For every integer $N \geq M$, split the average into the initial block and the tail block: \begin{align*} \left|\frac{1}{N+1}\sum_{n=0}^{N} e_n\right| &\leq \frac{1}{N+1}\sum_{n=0}^{N} |e_n| \\ &= \frac{1}{N+1}\sum_{n=0}^{M-1} |e_n| + \frac{1}{N+1}\sum_{n=M}^{N} |e_n|. \end{align*} The initial block is exactly $A/(N+1)$. On the tail block, every summand is smaller than $\varepsilon/2$, so \begin{align*} \frac{1}{N+1}\sum_{n=M}^{N} |e_n| &\leq \frac{1}{N+1}\sum_{n=M}^{N} \frac{\varepsilon}{2} \\ &= \frac{N-M+1}{N+1}\cdot \frac{\varepsilon}{2} \\ &\leq \frac{\varepsilon}{2}. \end{align*} Therefore \begin{align*} \left|\frac{1}{N+1}\sum_{n=0}^{N} e_n\right| \leq \frac{A}{N+1} + \frac{\varepsilon}{2}. \end{align*} Since $A$ is fixed, choose an integer $N_0 \geq M$ such that \begin{align*} \frac{A}{N_0+1} < \frac{\varepsilon}{2}. \end{align*} Then for every integer $N \geq N_0$, \begin{align*} \left|\frac{1}{N+1}\sum_{n=0}^{N} e_n\right| < \varepsilon. \end{align*} This proves that the Cesaro averages of the errors converge to $0$. [/guided] [/step] [step:Conclude that the Cesaro means converge to the original sum] From the preceding step, \begin{align*} \frac{1}{N+1}\sum_{n=0}^{N} e_n \to 0 \end{align*} as $N \to \infty$. Using the identity established in the first step, \begin{align*} \sigma_N - s = \frac{1}{N+1}\sum_{n=0}^{N} e_n, \end{align*} we obtain $\sigma_N - s \to 0$, and hence $\sigma_N \to s$. Therefore the series $\sum_{n=0}^{\infty} a_n$ is Cesaro summable to $s$. [/step]