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
Best Answer
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$, $n \to \infty$.