Hausdorff summation

Remarks

Example: for $\mu_n = \frac{1}{n+1} = \int_0^1 t^n \, dt$ one gets $\Delta^{n-k} \mu_k = \int_0^1 t^k (1 - t)^{n-k} \, dt = \frac{k! \, (n-k)!}{(n+1)!}$. So $h_{nk} = \frac{1}{n+1}$, and $t_n = \frac{s_0 + \cdots + s_n}{n+1}$ is the Cesàro mean. Similarly, $\mu_n = 2^{-n}$ gives $h_{nk} = \binom{n}{k} 2^{-n}$, the Euler $(E, 1)$ means.

Non-example of regularity: for $\mu_0 = 1$ and $\mu_n = 0$ for $n \ge 1$, one has $\Delta^{n-k} \mu_k = 0$ for $k \ge 1$ and $\Delta^n \mu_0 = 1$. So $t_n = s_0$ for all $n$, and the method sends every sequence to its first term.

Hausdorff showed that $H_\mu$ is regular if and only if $\mu_n = \int_0^1 t^n \, d\chi(t)$ for a function $\chi$ of bounded variation on $[0, 1]$ with $\chi(0) = 0$, $\chi(1) = 1$ and $\chi$ continuous at $0$. In the non-example, $\chi$ jumps from $0$ to $1$ at $0$, which violates the last condition.