What are the conditions on $\lambda \textrm{ and } \mu$ of $$ \sum_{n=0}^{n=\infty} \frac{\mu^{n}(2 \lambda + \mu)}{2^{n + 1} (\lambda + \mu)^{n + 1}} = 1 $$ where, $\lambda, \mu$ are positive parameters. $n \in \mathbb{N}$.
Background:
The term $\frac{\mu^{n}(2 \lambda + \mu)}{2^{n + 1} (\lambda + \mu)^{n + 1}}$ comes from a probability distribution, so I want to confirm that the serial sum is indeed $1$.
The special cases of $\lambda = \mu = 1$ and $\lambda = 1/2, \mu = 1$ are valid for the serial sum to be $1$. On the other hand, it diverges when $\mu = 2$.
This is a geometric series with a first term of $\dfrac{2\lambda+\mu}{2\lambda+2\mu}$ and a common ratio of $\dfrac{\mu}{2\lambda+2\mu}$.
Since $\lambda,\mu > 0$, the common ratio is between $0$ and $1$ exclusive.
Hence, the sum is $\dfrac{\dfrac{2\lambda+\mu}{2\lambda+2\mu}}{1-\dfrac{\mu}{2\lambda+2\mu}} = \dfrac{2\lambda+\mu}{2\lambda+\mu} = 1$, for any values $\lambda,\mu > 0$.