# Cumulative function of a given function multiplied by a Poisson distribution

Can anyone help me solve the following issue?

I have this expression: $$\tau_1 = \sum_{n=0}^\infty \bigg(\frac{e^{-\lambda}\lambda^n}{n!}\frac{C}{1+A(1-B)^n}\bigg)$$ where A ($A \approx 9$), B ($B < 1$) and C ($C \approx 4)$ are constants.

I need to expressed it as a function of $\lambda$ ($\lambda$ is a variable that is fixed for each value of $\tau_1$), and not of $n$.

As an example of what I mean, I had this other expression: $$\tau_2 = \sum_{n=0}^\infty \bigg(\frac{e^{-\lambda}\lambda^n}{n!} (1-(1-D)^n)\bigg)$$

where D is also a constant ($D < 1$). I was able to reduce $\tau_2$ to: $$\tau_2 = 1-e^{-D \lambda}$$ which is a function of $\lambda$, but doesn't depend on $n$ anymore.

I don't know how to proceed with $\tau_1$ expression.

Any ideas/suggestions/hints?

Notice that the first series can be written as

$$\tau_1 = Ce^{-\lambda} \sum_{k = 0}^{+\infty} \frac{\lambda^k}{k!} \frac{1}{1 + Ax^k}$$

Where $x = 1-B$.

The series doesn't have a closed form, unless $A = 0$. In that trivial case the sum is just the exponential $e^{\lambda}$ whence

$$\tau_1 = C$$

But $A = 0$ is just a very particular case.

We cannot say if $A >0$ or $A<0$, because it depends upon $B$, and we have no data.

Supposing $x\in\mathbb{R}$ and $\lambda > 0$, a close form is very unlikely, or there could exist but it will be a bitter monster.