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I'm trying to compute: $$\sum_{k=0}^{\infty} \frac{(-1)^{k} \Gamma\left(\frac{2k+n+m-1}{2}\right)}{k!} \, x^{2k}.$$

I thought of the expression of the binomial series $${\displaystyle (1+x)^{\alpha }=\sum _{n=0}^{\infty }{\binom {\alpha }{n}}x^{n}} \qquad (*)$$ where $${\displaystyle {\binom {\alpha }{n}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}} \quad (\text{the generalized binomial coefficients}).$$ i.e., \begin{align*} \sum_{k=0}^{\infty} \frac{(-1)^{k} \Gamma\left(\frac{2k+n+m-1}{2}\right)}{k!} \, x^{2k}&= \sum_{k=0}^{\infty} \Gamma\left(\frac{2k+n+m-1}{2}\right) \, \frac{(-x^{2})^k}{k!}\\ &= \sum_{k=0}^{\infty} \Gamma\left(k+\frac{n+m-1}{2}\right) \, \frac{(-x^{2})^k}{k!}. \end{align*} In other words: \begin{align*} \sum_{k=0}^{\infty} \Gamma\left(k+\alpha\right) \, \frac{t^k}{k!}=?? \end{align*} What do I have to do, in order to get to the expreesion $(*)$. Thank's

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    $\begingroup$ It's more convenient to put $n+m-1 = j$ and $t=-x^2$ and work with $j$ and $t$. $\endgroup$ Commented Nov 2, 2021 at 22:45
  • $\begingroup$ Does this answer your question? Infinite series with gamma function $\endgroup$ Commented Nov 2, 2021 at 22:54
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    $\begingroup$ @Botnakov, No. Because, I have $\Gamma(k+\alpha)$ not $\Gamma(\frac{k}{2}+\alpha)$. $\endgroup$
    – Z. Alfata
    Commented Nov 2, 2021 at 22:56
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    $\begingroup$ This one can help: math.stackexchange.com/q/2132666 Look at the series in the question. $\endgroup$
    – Gary
    Commented Nov 2, 2021 at 22:58
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    $\begingroup$ Use the fact that $\Gamma(k + \alpha) = (\alpha)_k \Gamma(\alpha)$, where $(\,\cdot\,)_k$ is the Pochhammer symbol, and your sum looks like $\Gamma(\alpha)\sum_0^\infty\, (\alpha)_k t^k /k!$. Then have a look at the function ${}_1F_0(\alpha, -, t)$ and its simplification. $\endgroup$ Commented Nov 2, 2021 at 23:17

1 Answer 1

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Suppose that $|t|<1$. We have \begin{eqnarray*} \Gamma(z) = \int_0 ^{\infty} x^{z-1} e^{-x} dx \Longrightarrow \sum_{n=0}^{\infty} \frac{ \Gamma(n+s)}{n!}t^n =\sum_{n=0}^{\infty} \int_0 ^{\infty} \frac{t^n x^{n+s-1} e^{-x}}{n!} dx \end{eqnarray*} Now invert the sum & integral \begin{eqnarray*} \int_0 ^{\infty} x^{s-1} \sum_{n=0}^{\infty} \frac{(tx)^{n} }{n!} e^{-x}dx =\int_0 ^{\infty} x^{s-1} e^{x(t-1)} dx \end{eqnarray*} Put $u = x(1-t)$. We have $$\sum_{n=0}^{\infty} \frac{ \Gamma(n+s)}{n!}t^n = \int_0 ^{\infty} \frac{u^{s-1}}{(1-t)^{s-1}} e^{-u} \frac{du}{1-t} = \frac{1}{(1-t)^s}\int_0^{\infty}u^{s-1}e^{-u}du = \frac{\Gamma(s)}{(1-t)^s}$$

Addition: if $|t| \ge 1$ and $s \ge 1$ then $\sum_{n=0}^{\infty} \frac{ \Gamma(n+s)}{n!}t^n$ diverges because $ \frac{ \Gamma(n+s)}{n!}t^n \not \to 0$, $n \to \infty$.

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