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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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  • $\begingroup$ Should the $\alpha^{i-1}$ in the end be $\alpha^{j-1}$? $\endgroup$
    – mickep
    Jan 2, 2015 at 21:12
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    $\begingroup$ I'm afraid this isn't the Pochhammer symbol, which should be written as $ \prod \limits_{j=1}^{i}(\alpha+j-1) $. $\endgroup$
    – Bernard
    Jan 2, 2015 at 21:18
  • $\begingroup$ You're right, my mistake. $\endgroup$
    – David Pfau
    Jan 2, 2015 at 21:40
  • $\begingroup$ 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. $\endgroup$
    – rrogers
    Mar 22, 2015 at 18:44

1 Answer 1

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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.

${\displaystyle \sum_{n=0}^{\infty}\left(\alpha\right)^{n}\left(-\pi\right)^{n}=\frac{E_{\alpha}\left(\frac{1}{\pi}\right)}{\pi\cdot e^{-\frac{1}{\pi}}}}$

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