= Solution
For an <M M 1 queue>, the first service lasts on average $1/\mu$. During it, the mean number of arrivals is $\lambda/\mu$, and each arrival initiates another statistically identical busy-period family. Thus, when $\lambda<\mu$,
$$
\mathbb EB
=\frac1\mu+\frac\lambda\mu\mathbb EB,
$$
so the <mean busy period of an M-M-1 queue> is
$$
\boxed{\mathbb EB_{M/M/1}=\frac1{\mu-\lambda}.}
$$
For $\lambda=\mu$ the busy period is almost surely finite but has infinite mean, while for $\lambda>\mu$ it is infinite with positive probability; in either case its expectation is infinite.
For an <M-M-infinity queue>, stationary occupancy is <Poisson distribution>[Poisson] with mean $\rho=\lambda/\mu$, so the long-run probability of an empty system is
$$
\pi_0=e^{-\rho}.
$$
The process alternates between idle periods and busy periods. An idle period waits for the next <Poisson process>[Poisson arrival], so its mean is $1/\lambda$. The <renewal-reward theorem>, with reward equal to time spent empty, gives
$$
e^{-\rho}
=\frac{1/\lambda}{1/\lambda+\mathbb EB}.
$$
Solving yields the <mean busy period of an M-M-infinity queue>:
$$
\boxed{
\mathbb EB_{M/M/\infty}
=\frac{e^{\lambda/\mu}-1}{\lambda}.}
$$
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