Euler summation

Remarks

For $q = 0$ the inner sum is $a_p$ alone, and $(E, 0)$ is ordinary convergence.

Example: take $a_k = (-1)^k (k + 1)$, that is $1 - 2 + 3 - 4 + \cdots$, and $q = 1$. The identities $\sum_{k=0}^{p} \binom{p}{k} (-1)^k = 0$ for $p \ge 1$ and $\sum_{k=0}^{p} \binom{p}{k} (-1)^k k = 0$ for $p \ge 2$ show that only $p = 0$ and $p = 1$ contribute. These give $\frac12 \cdot 1 + \frac14 \cdot (1 - 2) = \frac14$, so the series is $(E, 1)$ summable to $1/4$.

Non-example: for $1 + 1 + 1 + \cdots$ and $q = 1$, the inner sum is $2^p$, and every term of the new series is $\frac12$. That series diverges, so $1 + 1 + 1 + \cdots$ is not $(E, 1)$ summable.