How to compute the sum $\displaystyle\sum_{n=0}^{\infty}nP_{n}$ Following Problem is from probability theory:Define  $G(n),P(n)\ge 0,n\in\mathbb{N}$,and such
$$\begin{cases}G(n)=e^{-\lambda}\cdot\dfrac{\lambda^n}{n!},\lambda>0\\
\displaystyle\sum_{j=0}^{n}G(j)P(n-j)+(n+1)[P(n+1)-P(n)]=0,\forall n\in \mathbb{N} ,\\
 \displaystyle \sum_{n=0}^{\infty}P_{n}=1,\text{ i.e. the measure of entire  sample space i equal to one}
\end{cases}$$
Compute the sum( the expectation)$\displaystyle\sum_{n=0}^{\infty}nP_{n}$
 A: The equations provided lead to the following:
\begin{align}
\sum_{n=0}^{\infty} G_{n} \, t^{n} &= e^{- \lambda} \, \sum_{n=0}^{\infty} \frac{(\lambda \, t)^{n}}{n!} = e^{-\lambda \, (1 - t)}. 
\end{align}
Now,
\begin{align}
0 &= \sum_{n=0}^{\infty} (n+1) \left(P_{n+1} - P_{n} \right) \, t^{n} + \sum_{n=0}^{\infty} \, \sum_{k=0}^{n} G_{k} \, P_{n-k} \, t^{n} \\
&= \sum_{n=0}^{\infty} (n+1) \left(P_{n+1} - P_{n} \right) \, t^{n} + \sum_{n=0}^{\infty} \, \sum_{k=0}^{\infty} G_{k} \, P_{n} \, t^{n+k} \\
&= \frac{1}{t} \, \sum_{n=0}^{\infty} n P_{n} \, t^{n} - \sum_{n=0}^{\infty} n \, P_{n} \, t^{n} - P(t) + e^{-\lambda (1-t)} \, P(t) \\
&= \frac{1-t}{t} \, \sum_{n=0}^{\infty} n \, P_{n} \, t^{n} - (1 - e^{-\lambda(1-t)}) \, P(t) 
\end{align}
this becomes
\begin{align}
\frac{\sum_{n=0}^{\infty} n \, P_{n} \, t^{n}}{P(t)} = \frac{t \, \left( 1 - e^{-\lambda (1-t)}\right)}{1-t}
\end{align}
where $P(t) = \sum_{n=0}^{\infty} P_{n} \, t^{n}$. By taking the limit as $t \to 1$ the result
\begin{align}
\lim_{t \to 1} \left\{ \frac{\sum_{n=0}^{\infty} n \, P_{n} \, t^{n}}{\sum_{n=0}^{\infty} P_{n} \, t^{n}} \right\} = \lim_{t \to 1} \left\{ \frac{t \, \left( 1 - e^{-\lambda (1-t)}\right)}{1-t} \right\}
\end{align}
leads to
\begin{align}
\sum_{n=0}^{\infty} n \, P_{n} = \lambda
\end{align}
which is the expected result. 
