How do I prove that the following expected values are equivalent?

Question

I have been having difficulties with the following problem:

Show that if $f(x)$ is bounded and $X\sim\mathrm{Poiss}(\lambda)$ then $\mathbb{E}[\lambda f(X+1)]=\mathbb{E}[Xf(X)]$. Here $\mathrm{Poiss(\lambda)}$ denotes the usual Poisson distribution with pmf $p(k)=e^{-\lambda}\lambda^k/k!$ for $k\geq0$.

From what I could find online, it seems it has something to do with exchangeable pairs and/or Poisson approximation, but I am struggling to get started with the problem. How do I prove that the following expected values are equivalent? Thank you in advance!

Answer 1

Be aware that:

$$\sum_{k=0}^{\infty}\lambda f\left(k+1\right)\frac{\lambda^{k}}{k!}=\sum_{k=1}^{\infty}f\left(k\right)\frac{\lambda^{k}}{\left(k-1\right)!}=\sum_{k=1}^{\infty}kf\left(k\right)\frac{\lambda^{k}}{k!}=\sum_{k=0}^{\infty}kf\left(k\right)\frac{\lambda^{k}}{k!}$$