I have a Bessel function of the first kind, whose series expansion is:
$ J_{\alpha} (t) = \displaystyle\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{t}{2}\right)^{2m+\alpha}.$
In my specific application, variable $t$ can be decomposed as $t=\frac{xy}{z}$, yielding
$ J_{\alpha} \left(\frac{xy}{z}\right) = \displaystyle\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{xy}{2z}\right)^{2m+\alpha}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)$
I would like to know if it is possible rewrite the above expression in the form
$\Xi\left(\frac{y}{z}\right)J_{\alpha} (x)$.
My guess is to rewrite (1) as
$ J_{\alpha} \left(\frac{xy}{z}\right) = \displaystyle\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}\left(\frac{y}{z}\right)^{2m}\left(\frac{y}{z}\right)^{\alpha} = \left(\frac{y}{z}\right)^{\alpha}\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}\left(\frac{y}{z}\right)^{2m}$.
Moreover, under the assumption that $|\frac{y}{z}|<1$, I would like to know if it is possible to factor out the sum of the geometric series $\displaystyle\sum_{m=0}^{+\infty}\left(\frac{y^2}{z^2}\right)^m$ as
$ J_{\alpha} \left(\frac{xy}{z}\right) = (\frac{y}{z})^{\alpha}\left(\frac{1}{1-\frac{y^2}{z^2}}\right)\displaystyle\sum_{m=0}^{+\infty}\frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m+\alpha}$.
Is it correct ? I have doubts on the last step.