I have been struggling with a problem involving a Markov Chain. To solve it I need to figure out the following sum:$$\sum_{k=1}^{\infty}\frac{[\lambda(k-1)+\theta][\lambda(k-2)+\theta][\lambda(k-3)+\theta]\cdots[\lambda+\theta]\theta}{(k!)^2\mu^k}$$
Where, $\lambda=0.3$, $\theta=0.1$, $\mu=0.03$.
Cheers.
Considering $$S=\sum_{k=1}^{\infty}\frac{[\lambda(k-1)+\theta][\lambda(k-2)+\theta][\lambda(k-3)+\theta]...[\lambda+\theta]\theta}{(k!)^2\mu^k}$$ and using a CAS, the numerator in the summation is $$\theta\prod_{i=1}^{k-1} (\lambda i +\theta)=\theta \lambda ^{k-1} \left(\frac{\theta +\lambda }{\lambda }\right)_{k-1}$$ where appears the Pochhammer symbol.
So, the result is $$S=\, _1F_1\left(\frac{\theta }{\lambda };1;\frac{\lambda }{\mu }\right)-1$$ where appears the Kummer confluent hypergeometric function. It can also write $$S=L_{-\frac{\theta }{\lambda }}\left(\frac{\lambda }{\mu }\right)-1$$ where appears the Laguerre polynomial.
Using the numbers $\lambda=0.3$, $\theta=0.1$, $\mu=0.03$, we then obtain $$S=1864.59998928641$$
I hope this helps.