Why is $\sum_{k=0}^\infty e^{-\beta k} \frac{e^{-\mu t} (\mu t)^k}{k!} = \exp[-\mu t(1 - e^{-\beta})]$?

Question

It seems that

$$\sum_{k=0}^\infty \frac{e^{-\mu t} (\mu t)^k}{k!} = e^{-\mu t} e^{\mu t} = 1$$

which leaves $\sum_{k=0}^\infty e^{-\beta k}$.

However, I don't know if those expressions can be separated in that way.

Answer 1

No, that isn't a valid manipulation: it is not the case that $\sum_k a_k b_k = (\sum a_k)(\sum b_k)$, which is what you tried to do.

However, trying to recognize an exponential series is a good idea. Note that \begin{align*} \sum_{k=0}^\infty e^{-\beta k} \frac{e^{-\mu t} (\mu t)^k}{k!} &= e^{-\mu t} \sum_{k=0}^\infty e^{-\beta k} \frac{ (\mu t)^k}{k!} \\ &= e^{-\mu t} \sum_{k=0}^\infty \frac{ (e^{-\beta}\mu t)^k}{k!} \\ &= e^{-\mu t} e^{e^{-\beta}\mu t} = e^{-\mu t + e^{-\beta}\mu t}. \end{align*} (In the first line, it is valid to take the factor $e^{-\mu t}$ out of the sum, because the index of summation is $k$ and $e^{-\mu t}$ does not depend on $k$.)

Answer 2

$e^{\mu t}$ does not depend on the running index $k$ so it can be pulled out of the sum. So $\sum\limits_{k=0}^{\infty} e^{-\beta k}\frac {e^{-\mu t}(\mu t)^{k}} {k!}=e^{-\mu t} \sum\limits_{k=0}^{\infty}\frac {(e^{-\beta }\mu t)^{k}} {k!}$.