Interesting behavior of the expansion of $_1F_2(\alpha/2;3/2,\alpha/2+1;y^2/4)$ near $y=\infty$ When we use Mathematica 10.0 to expand generalized hypergeometric function $_1F_2(\alpha/2;3/2,1+\alpha/2;y^2/4)$ near $y=\infty$ with $\alpha$ a complex number, we obtain:
$${_1F_2}(\alpha/2;3/2,1+\alpha/2;y^2/4)$$
$$\approx e^{-y}\frac{\alpha}{2y}\left(1-\frac{(2-\alpha)}{y}+\frac{(2-\alpha)(3-\alpha)}{y^2}
\frac{(2-\alpha)(3-\alpha)(4-\alpha)}{y^3}+...\right)$$
$$+e^{y}\frac{\alpha}{2y}\left(1+\frac{(2-\alpha)}{y}+\frac{(2-\alpha)(3-\alpha)}{y^2}+\frac{(2-\alpha)(3-\alpha)(4-\alpha)}{y^3}+...\right)+\frac{\sqrt{\pi}\alpha(iy/2)^{-\alpha}\Gamma(\alpha/2)}{4\Gamma((3-\alpha)/2)}\tag{1}$$
As pointed out by @RobertIsrael in an answer to my previous question, the infinite series in the parenthesis are $_2F_0$ functions:
$${_1F_2}(\alpha/2;3/2,1+\alpha/2;y^2/4)\approx e^{-y}\frac{\alpha}{2y}{_2F_0}(1,2-\alpha;;-y^{-1})+e^{y}\frac{\alpha}{2y}{_2F_0}(1,2-\alpha;;y^{-1})+\frac{\sqrt{\pi}\alpha(iy/2)^{-\alpha}\Gamma(\alpha/2)}{4\Gamma((3-\alpha)/2)}\tag{2}$$
@RobertIsrael also pointed out that $_2F_0(1,2-\alpha;;x)$ should diverge for all $x\not = 0$.
My questions are:
(A) Is it possible to prove an identity like (2)?
(B) Since left hand side of (2) is $_1F_2$ function, so it is convergent. how to resolve the apparent divergent phenomenon on the right hand side of (2)?
(C) In (2), Where does the $\Gamma()$ part come from ?
Best-
mike
 A: (A) It IS possible to prove (1) and, by extension, (2), just lengthy. For the rest of this, let:
$$
F = {_1F_2}(\alpha/2;3/2,1+\alpha/2;y^2/4)
$$
Then, by Euler's Integral Transform for Hypergeometric Functions:
$$
F = \frac{\Gamma(\frac{3}{2})}{\Gamma(\frac{\alpha}{2})\Gamma(\frac{3-\alpha}{2})}\int_0^1 t^{\frac{\alpha}{2} - 1} (1 - t)^{\frac{1-\alpha}{2}}{_0F_1}(;1+\alpha/2;t \ y^2/4) \ dt
$$
Using the definition of $ _0F_1 $, we have:
$$
=\frac{\sqrt{\pi}}{2 \ \Gamma(\frac{\alpha}{2})\Gamma(\frac{3-\alpha}{2})} \int_0^1 t^{\frac{\alpha}{2} - 1} (1 - t)^{\frac{1-\alpha}{2}}\sum_{n \ = \ 0}^\infty \frac{(y^2 / 4)^n}{(1+\alpha / 2)_n n!}t^ndt
\\=\frac{\sqrt{\pi}}{2 \ \Gamma(\frac{\alpha}{2})\Gamma(\frac{3-\alpha}{2})} \sum_{n \ = \ 0}^\infty \frac{(y^2 / 4)^n}{(1+\alpha / 2)_n n!}\int_0^1 t^{n \ + \ \frac{\alpha}{2} - 1} (1 - t)^{\frac{1-\alpha}{2}}dt
$$
This integral is then just a form of the Beta Function, giving:
$$
=\frac{\sqrt{\pi}}{2 \ \Gamma(\frac{\alpha}{2})\Gamma(\frac{3-\alpha}{2})} \sum_{n \ = \ 0}^\infty \frac{(y^2 / 4)^n}{(1+\alpha / 2)_n n!} B \ (n + \frac{\alpha}{2},\frac{3-\alpha}{2})
\\ =\frac{\Gamma(1+\frac{\alpha}{ 2})\Gamma(\frac{3-\alpha}{2})\sqrt{\pi}}{2 \ \Gamma(\frac{\alpha}{2})\Gamma(\frac{3-\alpha}{2})} \sum_{n \ = \ 0}^\infty \frac{\Gamma(n + \frac{\alpha}{2})(y^2 / 4)^n}{\Gamma(n + \frac{\alpha}{2}+1)\Gamma(n + \frac{3}{2})n!}
\\ =\frac{\sqrt{\pi} \ \alpha}{4} \sum_{n \ = \ 0}^\infty \frac{(y^2 / 4)^n}{(n + \frac{\alpha}{2})\Gamma(n + \frac{3}{2})n!}
\\ =\frac{\sqrt{\pi} \ \alpha}{4} \sum_{n \ = \ 0}^\infty \frac{y^{2n}}{(2n + \alpha)\Gamma(n + \frac{3}{2})n!2^{2n-1}} \qquad \mathbf{(*)}
$$
Legendre's Duplication Formula for the Gamma function says:
$$
n! \ \Gamma(n+3/2) \ 2^{2n-1} = \frac{\sqrt{\pi} \ (2n+1)!}{4}
$$
Therefore:
$$
F=\frac{\alpha}{y} \sum_{n \ = \ 0}^\infty \frac{y^{2n+1}}{(2n + \alpha)(2n+1)!}
\\=\frac{\alpha}{2y} \sum_{n \ = \ 0}^\infty \frac{y^n - (-y)^n}{(n + \alpha - 1)n!}
\\=\frac{\alpha y^{-\alpha}}{2}\left(y^{\alpha - 1} \sum_{n \ = \ 0}^\infty \frac{(-1)^n (-y)^n}{(n + \alpha - 1)n!} - y^{\alpha - 1} \sum_{n \ = \ 0}^\infty \frac{(-1)^n y^n}{(n + \alpha - 1)n!} \right)
$$
Then, by the definition of the Lower and Upper Incomplete Gamma Functions:
$$
=\frac{\alpha y^{-\alpha}}{2}\left((-1)^{1 - \alpha} \gamma(\alpha - 1,-y) - \gamma(\alpha - 1,y) \right)
\\=\frac{\alpha y^{-\alpha}}{2}\left((-1)^{1 - \alpha} \Gamma(\alpha - 1) - \Gamma(\alpha - 1) + \Gamma(\alpha - 1,y) + (-1)^{-\alpha} \Gamma(\alpha - 1,-y)\right)
\\=\frac{\alpha y^{-\alpha}}{2} \Gamma(\alpha - 1,y) + \frac{\alpha (-y)^{-\alpha}}{2} \Gamma(\alpha - 1,-y) + \frac{\alpha y^{-\alpha} \Gamma(\alpha - 1)}{2}((-1)^{1 - \alpha} - 1)
$$
The Upper Incomplete Gamma Function has a well known asymptotic formula that results in F having the form of exactly (1) of your question, see the Wikipedia entry.  
(B) The radius of convergence for this asymptotic on the Incomplete Gamma is zero, meaning it diverges everywhere, except zero. This carries over for F, which explains why a 1F2 seems to diverge. However, I believe that if you only take a finite number of terms, it gives you any degree of accuracy you would like, it is only in taking all terms up to infinity that causes the divergence.  
(C) The Gamma term at the end in your derived formula is equivalent to the one that I fleshed out, just in a different form, more friendly to complex numbers. The Gammas are there simply because the Gammas are really everywhere. The Pochhammer symbols coming from the Hypergeometric formulae directly link to the Gamma function, thus causing it to be around often.  
There is probably a much better way to show this. This is just a way.
Edit: As stated in the comments, , by using the definition of a 1F2, one may reach (*) much more quickly:
$$
F = \sum_{n=0}^\infty \frac{(\alpha / 2)_n (y^2 / 4)^n}{(3 / 2)_n (1 + \alpha / 2)_n n!}
\\= \sum_{n=0}^\infty \frac{(\alpha / 2 + n - 1)! (\alpha / 2)! \Gamma(3 / 2) y^{2n}}{(\alpha / 2 - 1)! (\alpha / 2 + n)!\Gamma(3 / 2 + n) 2^{2n} n!}
\\= \frac{\sqrt{\pi} \ \alpha}{4} \sum_{n \ = \ 0}^\infty \frac{y^{2n}}{(2n + \alpha)\Gamma(n + \frac{3}{2})n!2^{2n-1}}
$$
Thus providing a shorter derivation.
