Cesàro summation

Remarks

For $\alpha = 0$ one has $\sigma^0_n = s_n$, which is ordinary convergence. For $\alpha = 1$ one has $A^1_m = m + 1$, and $\sigma^1_n = \frac{1}{n+1} \sum_{k=0}^{n} (n - k + 1) a_k = \sigma_n$, so clause 2 extends clause 1.

Example: for $1 - 1 + 1 - \cdots$ the partial sums are $1, 0, 1, 0, \dots$. So $\sigma_{2m} = \frac{m+1}{2m+1}$ and $\sigma_{2m+1} = \frac{1}{2}$, and both tend to $1/2$. The series is $(C, 1)$ summable to $1/2$.

Non-example: for $1 - 2 + 3 - 4 + \cdots$ the partial sums are $1, -1, 2, -2, 3, \dots$. Their sums are $1, 0, 2, 0, 3, 0, \dots$, so $\sigma_{2m} = \frac{m+1}{2m+1} \to \frac12$ and $\sigma_{2m+1} = 0$. The means oscillate, so the series is not $(C, 1)$ summable. It is $(C, 2)$ summable to $1/4$.

If a series is $(C, \alpha)$ summable to $s$, it is $(C, \beta)$ summable to $s$ for every $\beta > \alpha$.