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Does anyone know how to evaluate the series $$\sum_{k=0}^{\infty}\frac{x^k}{k!}\frac{1}{a+k}$$ where $a\in\mathbb{R}$ or if it simplifies things $a\in\mathbb{R_{\geq0}}$. This looks like the exponential series but I don't even know where to start.

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  • $\begingroup$ Consider the differential of $g(x)=x^a\cdot f(x)$. $\endgroup$
    – Berci
    Commented Mar 30, 2022 at 21:54
  • $\begingroup$ Could you elaborate? I don't know what to do... $\endgroup$ Commented Mar 30, 2022 at 21:59
  • $\begingroup$ I think you'll get $g'(x)=x^{a-1}\cdot e^x$, then you need to integrate it, though I'm not sure how if $a$ is not integer.. $\endgroup$
    – Berci
    Commented Mar 30, 2022 at 22:02

1 Answer 1

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$$\begin{split} \sum_{k=0}^{\infty}\frac{x^k}{k!}\frac{1}{a+k} &= \sum_{k=0}^{\infty}\frac{x^k}{k!}\int_0^1 t^{a+k-1}dt\\ &= \int_0^1\sum_{k=0}^{\infty}\frac{x^k}{k!}t^{a+k-1}dt\\ &= \int_0^1t^{a-1}e^{xt}dt \end{split}$$

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  • $\begingroup$ Good catch! Thank you. $\endgroup$ Commented Mar 30, 2022 at 22:30
  • $\begingroup$ It was actually more than confusing. It was a bad mistake. I hope this time I got this right. $\endgroup$ Commented Mar 30, 2022 at 22:39
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    $\begingroup$ "Famous last words"... ;-) $\endgroup$ Commented Mar 30, 2022 at 22:39

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