I'm evaluating the following function
$$\sum_{t=1}^{\infty}{e^{-b(t-1)}(1-e^{-b})(1-\rho)^t}$$
I attempted to solve it by equating it to
$$\sum_{t=1}^{\infty}{e^{-b(t-1)}}(1-e^{-b})\sum_{t=1}^{\infty}(1-\rho)^t$$ $$=\frac{e^b}{e^b-1}(1-e^{-b})\frac{1-\rho}{\rho}$$ $$=\frac{1-\rho}{\rho}$$
However, different results from the numerical calculation of the original summation function suggests that this is not a legal operation. Is that correct?
Your series is convergent as soon as $|1-\rho|<e^b$ and it is equal to $$e^b(1-e^{-b})\sum_{t=1}^{\infty}[e^{-b}(1-\rho)]^t=(e^b-1)\cdot \frac{e^{-b}(1-\rho)}{1-e^{-b}(1-\rho)}=\frac{(e^b-1)(1-\rho)}{e^b-(1-\rho)}.$$