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This sum correctly gives the Mittag-Leffler function:

Sum[z^k/Gamma[α*k + α], {k, 0, ∞}]

MittagLefflerE[α, α, z]

Simply factoring the argument of Gamma makes the sum fail:

Sum[z^k/Gamma[α*(k + 1)], {k, 0, ∞}]

$$ \sum_{k=0}^{\infty}\frac{z^k}{\text{Gamma}[(1+k)\alpha]} $$

What is happening? Why isn't this Expand/Factor transformation applied automatically by the pattern-matcher in order to effect the sum, as it is usually done for other sums, integrals, etc.?


update: reported & confirmed by WR.

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  • $\begingroup$ The second form did not match any of the built-in pattern rules. Sometimes you can help Mathematica by manipulating the expression. In this case, use ExpandAll, i.e., Sum[z^k/Gamma[\[Alpha]*(k + 1)] // ExpandAll, {k, 0, \[Infinity]}] $\endgroup$ – Bob Hanlon Apr 17 at 18:52
  • $\begingroup$ @BobHanlon that much is clear. What surprises me is how in this case (and only in this case as far as I remember working with Mathematica for many decades) the manipulations aren't done automagically. For all other sums, integrals, etc. I have a lot of freedom in how exactly I formulate the task; only for ML functions do I need to be that careful. So my question is: what is different about ML sums? Clarified the question, thanks. $\endgroup$ – Roman Apr 17 at 18:57
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    $\begingroup$ You may use Sum[z^k/Gamma[ \ [Alpha]*(k + 1)], {k, 0, [Infinity]}, Regularization -> "Abel"] (or any other method) to this end. Mma does not know that $\alpha$ is real. $\endgroup$ – user64494 Apr 17 at 19:02
  • $\begingroup$ Perhaps it has to do with the fact that Sum has the attribute HoldAll $\endgroup$ – Bob Hanlon Apr 17 at 19:06
  • $\begingroup$ @user64494 the conditioned sum Assuming[α > 0, Sum[z^k/Gamma[α*(k + 1)], {k, 0, ∞}]] doesn't work either. The Abel trick works though, thanks! $\endgroup$ – Roman Apr 17 at 19:09

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