A study on half-order differentiation of exponential function

Before you read this topic I advise you to read the following posts first.

The Binomial Theorem proof by exponentials

Fractional Calculus of Zero

Disproving D^{1/2}e^x=e^x with explanation

The following picture is showing graphically the relation between $e^x$ expansion terms and Reciprocal gamma function.
The plot is for $x=1,x=2,x=3$
Also it shows the differentiation and integration manner of $e^x$.

Click on the image to enlarge it

So if we move the terms of $e^x$ half step to the direction of differentiation, we will get the terms of half-integers order differentiation.
$$D_x^{1/2}(e^x)=\cdots +\frac{x^{-3/2}}{(-3/2)!}+\frac{x^{-1/2}}{(-1/2)!}+\frac{x^{1/2}}{(1/2)!}+\frac{x^{3/2}}{(3/2)!}+\cdots$$
(Note that when I use non-integer factorials (a)! I am indeed referring to Gamma function)
Now the question which I am trying to answer is:
 "Does this series converges? and how to deal with it in case of it diverges?"

I will use $e_{\alpha}^x$ as annotation for $D_x^{\alpha}(e^x)$
I recently proved this nice relation where $a,b \in \Bbb R$
$$e^a\cdot e_{\alpha}^b=e_{\alpha}^{a+b} \tag1$$
Which I can prove it by using the binomial formula from the previous post
$$\frac{(a+b)^{\alpha}}{\alpha!}=\sum_{k \in \Bbb Z} \frac{a^k}{k!} \cdot \frac{b^{\alpha-k}}{(\alpha-k)!}$$
And I already proved ,in the previous post, that the binomial theorem is actually a term of two multiplied exponential function ($\alpha$ can be any real number).

By using $eq(1)$ let us now test the relation for $\alpha=\frac12$
For $a=0$ we have,
$e^0\cdot e_{1/2}^b=e_{1/2}^{0+b} \Rightarrow e_{1/2}^b=e_{1/2}^b$
and that is a logical result.

For $b=0$ we have,
$e^a\cdot e_{1/2}^0=e_{1/2}^{a+0} \Rightarrow e^a \cdot e_{1/2}^0=e_{1/2}^a$
We all know that $e_{1/2}^0=\infty$
Hence $$e_{1/2}^a=\infty \cdot e^a$$

Therefore the expansion series of $e_{1/2}^a$ is not well-defined, but actually if we look to fractional calculus of 'zero number' which I studied it before in an earlier post makes the whole fractional calculus undefined and for that reason we always set the zero fractional calculus value to zero.

Thus we always have to eliminate the expansion series terms of $e_{1/2}^a$  that met the half-order differentiation of zero expansion so that it finally converges.
$$D_x^{1/2}(e^x)=\frac{x^{-1/2}}{(-1/2)!}+\frac{x^{1/2}}{(1/2)!}+\frac{x^{3/2}}{(3/2)!}+\frac{x^{5/2}}{(5/2)!}+\cdots$$