Mittag-Leffler summation

Remarks

For $a_k = z^k$ one gets $M(\alpha) = \sum_{k \ge 0} \frac{z^k}{\Gamma(1 + \alpha k)}$, the Mittag-Leffler function $E_\alpha(z)$.

Example: let $\sum a_k$ converge absolutely to $s$. Since $\Gamma(x) \ge \frac12$ for all $x \ge 1$, one has $\abs{a_k / \Gamma(1 + \alpha k)} \le 2 \abs{a_k}$ for all $\alpha$ and $k$. Each term tends to $a_k$ as $\alpha \to 0^+$, so by dominated convergence for series $M(\alpha) \to s$.

Non-example: for $1 + 1 + 1 + \cdots$, the terms $1/\Gamma(1 + \alpha k)$ are positive and tend to $1$ for each $k$. So for each $N$, $M(\alpha) \ge \sum_{k=0}^{N} \frac{1}{\Gamma(1 + \alpha k)} \to N + 1$ as $\alpha \to 0^+$, and $M(\alpha)$ has no finite limit.