Let $\mu_n = n\mu$ for $1 \le n \le 3$ and $\mu_n = 3\mu$ for $n \ge 4$. Let $\lambda_n = \lambda$ for all $n \in \mathbb{N}_0$. Define $\rho := \frac{\lambda}{\mu}$.
How would I expand $\displaystyle\sum\limits_{n=0}^\infty \prod_{j=0}^{n-1}\frac{\lambda_j}{\mu_{j+1}}$? Im confused because there is a summation and a product. I know that $\displaystyle\ \prod_{j=0}^{n-1}\frac{\lambda_j}{\mu_{j+1}} = \frac{\lambda_0}{\mu_1}.\frac{\lambda_1}{\mu_2}.\frac{\lambda_2}{\mu_3}.\frac{\lambda_3}{\mu_4}.\frac{\lambda_4}{\mu_5}.\frac{\lambda_5}{\mu_6} \dots = \frac{\lambda}{\mu}.\frac{\lambda}{2\mu}.\frac{\lambda}{3\mu}.\frac{\lambda}{3\mu}.\frac{\lambda}{3\mu}.\frac{\lambda}{3\mu}\dots = \rho.\frac{\rho}{2}.\frac{\rho}{3}.\frac{\rho}{3}.\frac{\rho}{3}.\frac{\rho}{3}\dots$
But how do I deal with the summation sign and expand $\displaystyle\sum\limits_{n=0}^\infty \prod_{j=0}^{n-1}\frac{\lambda_j}{\mu_{j+1}}$?
Edit: I checked the answer and it said $\displaystyle\sum\limits_{n=0}^\infty \prod_{j=0}^{n-1}\frac{\lambda_j}{\mu_{j+1}} = \displaystyle\sum\limits_{n=0}^{2}\frac{\rho ^n}{n!} + \frac{\rho ^3}{3!}\displaystyle\sum\limits_{n=0}^{\infty}\left(\frac{\rho}{3}\right)^n$ , but how would I get from the left side of the equality to the right?