How to show $\sum_0^\infty \frac{x\lambda^x} {x!} = \lambda e^\lambda$? I know that $\sum_0^\infty \frac{\lambda^x} {x!} = e^\lambda$, but I'm having a really difficult time dealing with the extra $x$.
 A: \begin{align}
\sum_{x=0}^\infty \frac{x\lambda^x}{x!} & = \sum_{x=1}^\infty \frac{x\lambda^x}{x!} & & \text{(since the first term is $0$)} \\[10pt]
& = \sum_{x=1}^\infty \frac{\lambda^x}{(x-1)!} & & \text{(since $\dfrac x {x!} = \dfrac 1 {(x-1)!}$)} \\[10pt] 
& = \lambda\sum_{x=1}^\infty \frac{\lambda^{x-1}}{(x-1)!} \\[10pt]
& = \lambda\sum_{y=0}^\infty \frac{\lambda^y}{y!} & & \text{where $y=x-1$} \\
& & & \text{As $x$ goes through $1,2,3,\ldots\,{}$,} \\
& & & \text{then $y$ goes through $0,1,2,3,\ldots\,{}$.}
\\[10pt]
& = \lambda\sum_{x=0}^\infty \frac{\lambda^x}{x!} =\cdots & & \text{(This is an “alphabetic} \\
& & & \phantom{(}\text{variant'' of the previous line.)}
\end{align}
A: The additional $x$ (from your initial knowledge) should lead you to think about derivatives with respect to $\lambda $. Remember a polynomial differentiates as:
$$ \left(x^n\right)' = nx^{n-1}$$. So you can rewrite in a close form:
$$ \sum_{x=0}^\infty \frac{x\lambda^x}{x!} = \lambda\sum_{x=0}^\infty \frac{x\lambda^{x-1}}{x!}. $$ If your present cursus includes convergent series and Fubini theorem, you can integrate the whole formula, switch $\sum$ and $\int$ signs, and get the desired result.
A: Very inelegantly and in schoolboy fashion we can write
\begin{align}
\sum_{x=0}^\infty x\frac{\lambda^x}{x!}
&= 0\frac{\lambda^0}{0!} + 1\frac{\lambda^1}{1!} + 2\frac{\lambda^2}{2!}
+ 3\frac{\lambda^3}{3!} + 4\frac{\lambda^4}{4!} + \cdots\\
&= 0 + \lambda +\frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!}+\cdots\\
&= \lambda +\frac{\lambda^2}{1!} + \frac{\lambda^3}{2!} + \frac{\lambda^4}{3!}+\cdots\\
&= \lambda\left[1 + \frac{\lambda^1}{1!} + \frac{\lambda^2}{2!}
+ \frac{\lambda^3}{3!} + \cdots\right]\\
&= \lambda e^\lambda
\end{align}
