Probability question using no-memory property of exponential distribution 
A customer must be served first by server 1, then by server 2, and
  finally by server 3. The amount of time required for service by server
  i is an exponential random variable with rate µi , for i = 1, 2, 3.
  Suppose you enter the system when it contains a single customer who is
  being served by server 3.

Find the expected amount of time that you spend in the system. (Whenever you encounter a busy server, you must wait for the service in progress to end before you can enter service.)
Attempt:
$1/\mu_1+1/\mu_2+1\mu_3 \frac{\mu_1}{\mu_1+\mu_2} \frac{\mu_2}{\mu_2+\mu_3)}1/\mu_3$
 A: Let $T$ be the amount of time you take to clear the system; $T_n$ be the amount of time you spend in an empty cue $n$; and $S_3$ be the amount of time the other customer spends in cue 3.   Clearly the critical issue is whether the other customer has cleared the cue 3 by the time you reach it.
$$\begin{align}\mathsf E[T] =~& \mathsf E[~\max(S_3, T_1+T_2)~]+ \mathsf E[~T_3~]
\\[1ex] = ~& \mathsf E[~\max(S_3, T_1+T_2)~]+ \dfrac 1{\mu_3}
\\[1ex] \ddots &
\end{align}$$

Now $$\begin{align}
\mathsf E[~\max(S_3,T_1+T_2)~] ~=~& \int_0^\infty \left(t\int_0^t f_{S_3}(s)\operatorname d s+\int_t^\infty sf_{S_3}(s)\operatorname d s \right) f_{T_1+T_2}(t)\operatorname d t
\\ =~& \int_0^\infty \left(t\int_0^t \mu_3\mathsf e^{-\mu_3s}\operatorname d s+\int_t^\infty s\mu_3\mathsf e^{-\mu_3s}\operatorname d s \right) f_{T_1+T_2}(t)\operatorname d t
\\ =~& \int_0^\infty \left(t(1-\mathsf e^{-\mu_3 t})+\frac{1+\mu_3t}{\mu_3}\mathsf e^{-\mu_3 t}+t \right) f_{T_1+T_2}(t)\operatorname d t
\\ =~& \int_0^\infty \left(\tfrac 1{\mu_3}\mathsf e^{-\mu_3 t}+t \right) f_{T_1+T_2}(t)\operatorname d t
\end{align}$$
Can you continue from here?
A: I'm not sure if this is correct, but you can do this with a conditioning argument: $E(T_1+T_2+T_3)+E(S|S>T_1+T_2)P(S>T_1+T_2)+E(S|S<T_1+T_2)P(S<T_1+T_2)=\frac{1}{\mu_1}+\frac{1}{\mu_2}+\frac{1}{\mu_3}+\frac{\mu_1}{\mu_1+\mu_3}\frac{\mu_2}{\mu_2+\mu_3}\frac{1}{\mu_3}$, since $E(S|S<T_1+T_2)=0$.
