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
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...)