Is there any close form of
$$\sum_{k=0}^{\infty} C_{m+k}^{k} \Lambda^{k}$$
where $C_{m+k}^{k}=\frac{(m+k)!}{k!\cdot m!}$.
$m$ is a given integer and $\Lambda\in (0,1)$ is also a given constant.
I know this converges but just cannot work out a better form.
Mathematica knows the answer: $$ \sum_{k=0}^\infty \frac{(m+k)!}{m!\,k!} x^k = (1-x)^{-(m+1)} $$ A proof can be constructed easily with the Euler representation of the gamma function, $$ (m+k)! = \int_0^\infty t^{m+k} e^{-t} dt $$ Insert the last equation, switch sum and integral, do the sum as an exponential, perform the integral explicitly using the previous formula with $k=0.$ The interchange is justified since $0<x<1.$