$ \sum_{y=1}^\infty {}_1F_1(1-y;2;-\pi\lambda c) \frac{\lambda^y}{y!} $ I am not able to solve the following sum. Can you please provide any hints ?
$$ \sum_{y=1}^\infty {}_1F_1(1-y;2;-\pi\lambda c) \frac{\lambda^y}{y!} $$
Note that the 3rd parameter of the Confluent Hypergeometric Function includes parameter λ.
Thank you for your time
 A: Integral method using the associated Laguerre polynomials with $\;x:=-\pi\lambda c$
\begin{align}
\sum_{n=1}^\infty {}_1F_1(1-n;2;x) \frac{\lambda^n}{n!}&=\sum_{n=1}^\infty \left[1+\sum_{k=1}^\infty \frac{(1-n)_k}{(2)_k}\frac{x^k}{k!}\right] \frac{\lambda^n}{n!}\\
&=\sum_{n=1}^\infty \sum_{k=0}^{n-1} \frac{(n-1)!}{(n-k-1)!(k+1)!}\frac{(-x)^k}{k!} \frac{\lambda^n}{n!}\\
&=\sum_{m=0}^\infty \sum_{k=0}^m \frac{m!}{(m-k)!(k+1)!}\frac{(-x)^k}{k!} \frac{\lambda^{m+1}}{(m+1)!}\quad\text{for}\ m:=n-1\\
&=\sum_{m=0}^\infty \frac{ L_m^1(x)}{(m+1)!}\frac{\lambda^{m+1}}{m+1}\\
&=\int \sum_{m=0}^\infty \frac{ L_m^1(x)}{(m+1)!}\lambda^m\;d\lambda\\
&=\int \frac{e^{\lambda}J_1\bigl(2\sqrt{x\lambda}\bigr)}{\sqrt{x\lambda}}\;d\lambda\\
&=\int_0^{2\sqrt{x\lambda}} \frac{e^{u^2/(4x)}}x\;J_1(u)\;du\quad\quad\text{(using}\ u:=2\sqrt{x\lambda}\,\text{)}\\
\end{align}
Since $\ \displaystyle L_m^1(x)= \sum_{k=0}^m \frac{(m+1)!}{(m-k)!\,(1+k)!\,k!}(-x)^k\ $ and $\ \displaystyle \sum_{m=0}^\infty \frac{L_m^1(x)}{(m+1)!}\lambda^m=\frac{e^{\lambda}J_1\bigl(2\sqrt{x\lambda}\bigr)}{\sqrt{x\lambda}}$
(formula $(5)$ and $(18)$ from the MathWorld link)

Initial method (returning alternative formulations...)
Let's expand the hypergeometric series using the Pochhammer symbols :
\begin{align}
\sum_{n=1}^\infty {}_1F_1(1-n;2;-\pi\lambda c) \frac{\lambda^n}{n!}&=\sum_{n=1}^\infty \left[1+\sum_{k=1}^\infty \frac{(1-n)_k}{(2)_k}\frac{(-\pi\lambda c)^k}{k!}\right] \frac{\lambda^n}{n!}\\
&=e^\lambda-1+\sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{(n-1)!}{(n-k-1)!(k+1)!}\frac{(\pi\lambda c)^k}{k!} \frac{\lambda^n}{n!}\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \sum_{n=k+1}^\infty\frac{\lambda^n}{n\;(n-k-1)!}\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \sum_{n=k+1}^\infty\frac{\lambda^{n-1}}{(n-k-1)!}d\lambda\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \lambda^k\sum_{m=0}^\infty\frac{\lambda^m}{m!}d\lambda\\
&=e^\lambda-1+ \sum_{k=1}^\infty \frac{(\pi\lambda c)^k}{k!(k+1)!} \int \lambda^k e^{\lambda}\;d\lambda\\
&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\gamma(k+1,-\lambda)}{k!(k+1)!}\\
\end{align}
(using the definition of the incomplete gamma function :
$\ \displaystyle\gamma(a,x)=\int_0^xt^{a-1}e^{-t}\;dt\ $)
\begin{align}
\sum_{n=1}^\infty {}_1F_1(1-n;2;-\pi\lambda c) \frac{\lambda^n}{n!}&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;k!\left(1-e^{\lambda}\sum_{m=0}^k\frac {(-\lambda)^m}{m!}\right)}{k!(k+1)!}\\
&=e^\lambda-1-\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\left(1-e^{\lambda}\sum_{m=0}^k\frac {(-\lambda)^m}{m!}\right)}{(k+1)!}\\
&=e^\lambda+\frac{e^{-\pi\lambda c}-1}{\pi\lambda c}+e^{\lambda}\sum_{k=1}^\infty \frac{(-\pi\lambda c)^k\;\sum_{m=0}^k\frac {(-\lambda)^m}{m!}}{(k+1)!}\\
\end{align}
At this point it would be interesting to know if :
$$f(x,y):=\sum_{k=0}^\infty \frac{x^k}{k!}\;\sum_{m=0}^k\frac {y^m}{m!}$$
may be simplified (I don't know...)
A: $\sum\limits_{y=1}^\infty{}_1F_1(1-y;2;-\pi\lambda c)\dfrac{\lambda^y}{y!}$
$=\sum\limits_{y=0}^\infty{}_1F_1(-y;2;-\pi\lambda c)\dfrac{\lambda^{y+1}}{(y+1)!}$
$=\sum\limits_{y=0}^\infty\sum\limits_{n=0}^y\dfrac{\pi^nc^n\lambda^{n+y+1}}{(2)_nn!(y-n)!(y+1)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{y=n}^\infty\dfrac{\pi^nc^n\lambda^{n+y+1}}{(2)_nn!(y-n)!(y+1)}$
$=\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{\pi^nc^n\lambda^{2n+y+1}}{(2)_nn!y!(n+y+1)}$
$=\int_0^\lambda\sum\limits_{n=0}^\infty\sum\limits_{y=0}^\infty\dfrac{\pi^nc^n\lambda^nx^{n+y}}{(2)_nn!y!}~dx$
$=\int_0^\lambda\sum\limits_{n=0}^\infty\dfrac{\pi^nc^n\lambda^nx^ne^x}{(2)_nn!}~dx$
$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}\pi^nc^n\lambda^nx^ke^x}{(2)_nk!}\right]_0^\lambda$ (according to http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n-k}\pi^nc^n\lambda^{n+k}e^\lambda}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^nc^n\lambda^n}{(2)_n}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^{n-k}\pi^nc^n\lambda^{n+k}e^\lambda}{(2)_nk!}-\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^nc^n\lambda^n}{(n+1)!}$
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^n\pi^{n+k}c^{n+k}\lambda^{n+2k}e^\lambda}{(2)_{n+k}k!}-\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}\pi^{n-1}c^{n-1}\lambda^{n-1}}{n!}$
$=e^\lambda\Phi_3(1,2;-\pi c\lambda,\pi c\lambda^2)+\dfrac{e^{-\pi c\lambda}-1}{\pi c\lambda}$ (according to http://en.wikipedia.org/wiki/Humbert_series)
