I can't understand this difference equation step I am working on birth-death processes and I can't understand a step that is taken in a proof.
The mean of a process is defined as
$$\mu(t) = \sum_{n=1}^{\infty}np_n(t)$$
At certain stage in the proof we go from 
$$\lambda \sum_{n=0}^{\infty}(n+1)p_n(t) + \lambda \sum_{n=1}^{\infty}np_n(t)$$
to
$$=\lambda \mu(t) + \lambda\mu(t)$$
I don't see how this is valid. The expansions for the summations aren't the same. For instance, $1 \times p_0 + 2 \times p_1 + ... \ne 1 \times p_1 + 2 \times p_2 + ...$
Am I missing something here, anyone know what's going on?
Here is the proof


Basically, the definition of expected value gives $$\sum_{n=0}^\infty n p_n(t)=\vec{\mu}(t),$$ and the definition of a probability distribution gives $$\sum_{n=0}^\infty p_n(t)=1.$$
In particular, $$\sum_{n=0}^\infty (n+1)p_n(t) = \vec{\mu}(t)+1.$$

Being explicit:
\begin{align*}
&\phantom{{}={}} \lambda \sum_{n = 0}^\infty(n+1)p_n(t) - \lambda \vec{\mu}(t) - \mu \sum_{n=1}^\infty n^2 p_n(t) + \mu \sum_{n=2}^\infty n(n-1) p_n(t)\\
&= \lambda (\vec{\mu}(t)+1) - \lambda \vec{\mu}(t) - \mu \sum_{n=1}^\infty n^2 p_n(t) + \mu \sum_{n=1}^\infty n(n-1) p_n(t) & \text{adding a zero addend}\\
&= \lambda - \mu \sum_{n=1}^\infty (n(n-1)-n^2) p_n(t)\\
&= \lambda +\mu \sum_{n=1}^\infty n p_n(t)\\
&= \lambda+ \mu \sum_{n=0}^\infty n p_n(t) & \text{adding a zero addend}\\
&= \lambda+\mu\vec{\mu}(t).
\end{align*}
