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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?

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Isn't it just a geometric series? –  Gerry Myerson Apr 29 '12 at 23:28
    
@GerryMyerson Sort of I guess, I just edited my post to add some details, maybe it might help? –  Richard Apr 29 '12 at 23:48
    
In your edit, $k$ appears on the right but not on the left, so something's wrong. –  Gerry Myerson Apr 30 '12 at 1:44
    
@GerryMyerson opps sorry, I forgot to replace $k$ with $3$. In this case I was letting $k =3$. I fixed it now. Any ideas? –  Richard Apr 30 '12 at 12:07
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1 Answer 1

up vote 2 down vote accepted

When $n=0$, the product is empty (it goes $j$ from $0$ to $-1$); by convention, empty products are $1$.

When $n=1$, the product is $\lambda_0/\mu_1=\lambda/\mu=\rho$.

When $n=2$, the product is $(\lambda/\mu)(\lambda/2\mu)=\rho^2/2$.

So, the first three terms are $1+\rho+(1/2)\rho^2$, and that's your $\sum_{n=0}^2(\rho^n/n!)$.

For $n=3,4,5,\dots$, each term in the sum is $\lambda_n/\mu_{n+1}=\lambda/3\mu=\rho/3$ times the previous term, so you have a geometric series with first term $\rho^3/3!$ and common ratio $\rho/3$. There's your $(\rho^3/3!)\sum_{n=0}^{\infty}(\rho/3)^n$.

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Thanks @GerryMyerson , understood fully and perfect answer :) Cheers again. –  Richard Apr 30 '12 at 13:53
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