Sum involving the hypergeometric function, power and factorial functions I am finding some trouble in calculating the following sum involving the hypergeometric function, power and factorial functions. 
$$
\sum_{y=1}^\infty \mathrm{e}^z \cdot {}_1F_1\left(1-y;2;-z\right) \frac{\mu^y}{y!} \mathrm{e}^{-\mu} 
$$
Could you please provide me some hints to solve this sum?
Thank you in advance
 A: *

*Multiply your sum by $e^{\mu-z}$ and differentiate with respect to $\mu$ to get
$$\sum_{n=0}^{\infty} {}_1F_1(-n,2,-z)\frac{\mu^n}{n!}\tag{1}$$

*Use the (I hope, accidentally) unaccepted answers to your question here to express $_1F_1$ in the sum as generalized Laguerre polynomials:
$$ \frac{_1F_1(-n,2,-z)}{n!}=\frac{L_n^{(1)}(-z)}{(1+1)_n}.\tag{2}$$

*Use (2) and the generating function of Laguerre polynomials (formula 18.12.14) to sum up the series (1) to 
$$ \left(-\mu z\right)^{-1/2}e^{\mu}J_1(2\sqrt{-\mu z})=\left(\mu z\right)^{-1/2}e^{\mu}I_1(2\sqrt{\mu z})$$

*Integrate back with respect to $\mu$ and multiply the result by  $e^{z-\mu}$. The final result is
$$ e^{z-\mu}\int_0^{\mu}\left(\nu z\right)^{-1/2}e^{\nu}I_1(2\sqrt{\nu z})\,d\nu=z^{-1}e^{z-\mu}\int_0^{2\sqrt{\mu z}}e^{{x^2}/{4z}}I_1(x)\,dx.$$
I don't think the last integral can be expressed in terms of elementary or reasonably simple special functions.

A: $\sum\limits_{y=1}^\infty e^z\cdot{}_1F_1(1-y;2;-z)\dfrac{\mu^y}{y!}e^{-\mu}$
$=e^{z-\mu}\sum\limits_{y=0}^\infty{}_1F_1(-y;2;-z)\dfrac{\mu^{y+1}}{(y+1)!}$
$=e^{z-\mu}\sum\limits_{y=0}^\infty\sum\limits_{n=0}^y\dfrac{\mu^{y+1}z^n}{(2)_nn!(y-n)!(y+1)}$
$=e^{z-\mu}\sum\limits_{n=0}^\infty\sum\limits_{y=n}^\infty\dfrac{\mu^{y+1}z^n}{(2)_nn!(y-n)!(y+1)}$
$=e^{z-\mu}\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{\mu^{n+y+1}z^n}{(2)_nn!y!(n+y+1)}$
$=e^{z-\mu}\int_0^\mu\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{x^{n+y}z^n}{(2)_nn!y!}~dx$
$=e^{z-\mu}\int_0^\mu\sum\limits_{n=0}^\infty\dfrac{x^ne^xz^n}{(2)_nn!}~dx$
$=e^{z-\mu}\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}x^ke^xz^n}{(2)_nk!}\right]_0^\mu$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}\mu^kz^ne^z}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nz^ne^{z-\mu}}{(2)_n}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^{n-k}\mu^kz^ne^z}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^nz^ne^{z-\mu}}{(n+1)!}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\mu^kz^{n+k}e^z}{(2)_{n+k}k!}-\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}z^{n-1}e^{z-\mu}}{n!}$
$=e^z\Phi_3(1,2;-z,\mu z)+\dfrac{e^{-\mu}-e^{z-\mu}}{z}$ (according to http://en.wikipedia.org/wiki/Humbert_series)
