<s>[INST] Investigation of $\sum \limits_{k=-\infty}^\infty \frac{x^{k+n}}{ \Gamma(k+n+1)}$ where $n \in C$? $$e^x=\sum \limits_{k=0}^\infty \frac{x^k}{k!}$$
We can rewrite the equation as
$$e^x=\sum \limits_{k=0}^\infty \frac{x^k}{ \Gamma(k+1)} \tag{1}$$
because $x!=\Gamma(x+1)$  where $x$ is non-negative integer.
$\Gamma(x)$ (Gamma function) also has undefined property of any of the integers {0, −1, −2,... . }  at those points; it is a meromorphic function with simple poles at the non-positive integers.
$m$ is non-positive, $$\frac{1}{\Gamma(m)}=0 \tag{2}$$
Let's combine the property $(1)$ with $(2)$ and  the formula $(1)$ can be extended for $x \neq 0$
$$e^x=\sum \limits_{k=-\infty}^\infty \frac{x^k}{ \Gamma(k+1)} \tag{3}$$
If we change sum variable  $k'=k+n$ where n is an integer . We get
$$e^x=\sum \limits_{k'=-\infty}^\infty \frac{x^{k'+n}}{ \Gamma(k'+n+1)} \tag{4}$$
$n \in Z $ {...,-2,-1,0,1,2,...}
I wanted to  find  out if the equation (4) holds true if we select $n \in C $ for $x \neq 0$ ?
If we select , $n=\frac{1}{2}$
$$\Gamma(1/2)=\sqrt{\pi}$$
$$f(x)=\sqrt{x}\sum \limits_{k=-\infty}^\infty \frac{x^{k}}{ \Gamma(k+3/2)} \tag{5}$$
I have noticed $f(x)$ has the same derivative result as $e^x$.
$$f'(x)=f(x) \tag{6}$$
Proof:
$$f(x)=\sqrt{x}\sum \limits_{k=-\infty}^\infty \frac{x^{k}}{ \Gamma(k+3/2)} \tag{7}$$
$$f(x)=\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)+\frac{1}{\sqrt{\pi x}}(1-\frac{1}{2x}+\frac{3}{2^2x^2}-\frac{3.5}{2^3x^3}+.... ) \tag{8}$$
$$f'(x)=\frac{1}{\sqrt{\pi x}}+\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)-\frac{1}{\sqrt{\pi x}}(\frac{1}{2x}-\frac{3}{2^2x^2}+\frac{3.5}{2^3x^3}-.... ) \tag{9}$$
$$f'(x)=\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)+\frac{1}{\sqrt{\pi x}}-\frac{1}{\sqrt{\pi x}}(\frac{1}{2x}-\frac{3}{2^2x^2}+\frac{3.5}{2^3x^3}-.... ) \tag{10}$$
$$f'(x)=\frac{2\sqrt{x}}{\sqrt{\pi}}(1+\frac{2x}{3}+\frac{2^2x^2}{3.5}+\frac{2^3x^3}{3.5.7}+....)+\frac{1}{\sqrt{\pi x}}(1-\frac{1}{2x}+\frac{3}{2^2x^2}-\frac{3.5}{2^3x^3}+.... ) \tag{11}$$
Thus
$$f'(x)=f(x) \tag{12}$$
More generally,
$$f_n(x)=\sum \limits_{k=-\infty}^\infty \frac{x^{k+n}}{ \Gamma(k+n+1)} \tag{13}$$
where $n \in C $ for $x \neq 0$ ?
The function $f_n(x)$ has same differential property $f'_n(x)=f_n(x)$  with $e^x$.
Please advice how to prove or disprove that $f_n(x)=e^x$  where $n \in C $ for $x \neq 0$ ?
Thanks a lot
 [/INST] You might like to look at asymptotic expansions--a rich, fascinating field. See this excerpt for a quick intro and the book by Dingle, available courtesy of Michael Berry, who has himself written many fine articles on the subject, e.g., this one. (Wikipedia has a short bibliography.)
Coincidentally, just last night I was looking at the asymptotic expansion of the upper incomplete gamma function (being related to some special Sheffer polynomial sequences):
$$\frac{x^s}{(s-1)!} \Gamma(x,s)=x^s \int_1^{\infty} e^{-xt} \frac{t^{s-1}}{(s-1)!}dt \sim e^{-x}\sum_{n=1}^{\infty}  \frac{x^{s-n}}{(s-n)!}.$$
Taking the first four terms in the summation for $x=1.9$ and $s=2.8$ gives $.653$ in good agreement with the exact value $.657$, so we have a summation for the lower asymptotic series in terms of the incomplete gamma function--the Cinderella function of Tricomi.  (Sum initialization corrected 1/8/15.)
The full Borel-Laplace transform (Whitttaker and Watson) by interchange of the Taylor series summation for $e^{xt}$ and the integrations then is 
$$e^x=e^xx^s [\int_0^{1}+\int_1^{\infty}] e^{-xt} \frac{t^{s-1}}{(s-1)!}dt \sim \sum_{n=0}^{\infty}  \frac{x^{s+n}}{(s+n)!}+\sum_{n=1}^{\infty}  \frac{x^{s-n}}{(s-n)!}.$$
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