Absolute summability

Remarks

If $\sum_{n \ge 1} \abs{t_n - t_{n-1}}$ is finite, then $(t_n)$ is a Cauchy sequence, so it converges. Thus an absolutely $A$-summable series is $A$-summable.

Example: for the identity matrix, $t_n = s_n$ and $t_n - t_{n-1} = a_n$. So absolute summability by the identity method is absolute convergence.

Non-example: Grandi's series $1 - 1 + 1 - \cdots$ is Cesàro summable to $1/2$, but not absolutely Cesàro summable. Its partial sums are $1, 0, 1, 0, \dots$, so the Cesàro means are $t_{2m} = (m + 1)/(2m + 1)$ and $t_{2m+1} = 1/2$. Then $\abs{t_{2m} - t_{2m-1}} = 1/(2(2m + 1))$, and the sum of these numbers diverges.

For the Cesàro method the notation is $\abs{C, 1}$. For Abel summation one says that $\sum a_n$ is absolutely Abel summable when $x \mapsto \sum a_n x^n$ has bounded variation on $[0, 1)$.