Prove an equation holds in series: $\sum\limits_{x=0}^{\infty} xe^{-\lambda}\frac{\lambda^x}{x!}=\lambda$

Question

Let $\lambda\in \mathbb{R}$ be a constant. Prove $\sum\limits_{x=0}^{\infty} xe^{-\lambda}\frac{\lambda^x}{x!}=\lambda$.

I have already proved $\sum\limits_{x=0}^{\infty}e^{-\lambda}\frac{\lambda^x}{x!}=1$ and I can use Maclaurin $e^{ty}=\sum\limits_{x=0}^{\infty}\frac{(ty)^x}{x!}$.

Thank you all.

Answer 1

Hint: $$\sum\limits_{x=0}^{\infty} xe^{-\lambda}\frac{\lambda^x}{x!}=\lambda e^{-\lambda}\sum\limits_{x=1}^{\infty} \frac{\lambda^{x-1}}{\left ( x-1 \right )!}$$

Answer 2

As we know : $\displaystyle e^{kx} = \sum \frac{(kx)^n}{n!}$, so $\displaystyle ke^{kx} = \sum n \frac{(kx)^{n-1}}{n!}$, and we get : $\displaystyle k^2xe^{kx} = \sum n\frac{(kx)^n}{n!}$ and now evaluate when $k = 1, x = \lambda$