Is there a closed-form expression for the infinite series
$\sum_{i=0}^\infty (-\pi)^i\alpha^{(i)}$
For known $\pi,\alpha\in [0,1)$ where $\alpha^{(i)}$ is the rising factorial or Pochhammer symbol $\prod_{j=1}^i (\alpha+j-1)$
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Should the $\alpha^{i-1}$ in the end be $\alpha^{j-1}$? – mickep Jan 02 '15 at 21:12
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1I'm afraid this isn't the Pochhammer symbol, which should be written as $ \prod \limits_{j=1}^{i}(\alpha+j-1) $. – Bernard Jan 02 '15 at 21:18
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You're right, my mistake. – David Pfau Jan 02 '15 at 21:40
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I posted an answer without proof; hope that's okay. I understand using Borel's transform will also work for conversions of EGF to OGF. – rrogers Mar 22 '15 at 18:44
1 Answers
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From : http://www.mi.sanu.ac.rs/~gvm/radovi/AP-JIS2.pdf Eq: 6
$\sum_{n=0}^{\infty}\left(b\right)^{n}x^{n}=-\frac{E_{b}\left(-\frac{1}{x}\right)}{x\cdot e^{\frac{1}{x}}}$
i.e.
${\displaystyle \sum_{n=0}^{\infty}\left(\alpha\right)^{n}\left(-\pi\right)^{n}=\frac{E_{\alpha}\left(\frac{1}{\pi}\right)}{\pi\cdot e^{-\frac{1}{\pi}}}}$
rrogers
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