Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$ I am trying to compute the Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$. When I compute the Cesàro means, I get the following sequence
$$\left(1, 1, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \frac{6}{9}, \frac{7}{10}, \frac{8}{11}, \frac{8}{12}, \cdots\right)$$
Where does this sequence converge to? Is it $\frac{2}{3}$? I fail to see the pattern of this sequence. If I can just rewrite the sequence into a more general form, then I might be able to compute the limit.

Let $a_n = (-1)^n$ for $n\ge 0$. The partial sum $$s_k = \sum_{n = 0}^k a_n$$ for $k\ge 0$ is $1 , 0 , 1 , 0,\dots$ which clearly shows $s_k$ is divergent. Let $$t_n = \frac{1}{n}\sum_{k = 0}^{n-1}s_n$$for $n\ge 1$. So we have $t_1 = \frac{1}{1} , t_2 = \frac{1}{2} , t_3 = \frac{2}{3} , t_4 = \frac{2}{4} , t_5 = \frac{3}{5} , t_6 = \frac{3}{6} , \dots$ and we can easily see the pattern $t_n = \frac{\lfloor\frac{n+1}{2}\rfloor}{n}$. The question is $\lim_{n \to \infty} t_n$ . We know that $x \ge \lfloor x\rfloor \gt x - 1$, so we have $$\frac{n+1}{2} \ge \lfloor \frac{n+1}{2}\rfloor \gt \frac{n+1}{2} - 1 \implies \frac{n+1}{2n} \ge \frac{\lfloor \frac{n+1}{2}\rfloor}{n} \gt \frac{n+1}{2n} - \frac{1}{n}$$ Because $\lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \frac{n+1}{2n} - \frac{1}{n} = \frac{1}{2}$, according to the squeeze theorem we have $$\lim_{n \to \infty} t_n = \frac{1}{2}$$
In the same manner you can construct your sequence and compute the limit. The answer is

 $b_n = \frac{\lfloor\frac{2n+2}{3}\rfloor}{n} \implies \lim_{n \to \infty} b_n = \frac{2}{3}$

