Expected value $\frac{1}{x+1}$ of Poisson distribution

Question

If the random variable $X$ has a Poisson distribution with mean $\lambda$, derive an expression for the expected value $\frac{1}{X+1}$

My attempt, \begin{align*} \mathbb{E}\left[\frac{1}{X+1}\right] &=\sum_{x=0}^{\infty}\frac{1}{x+1}P(X=x) \\ & =\sum_{x=0}^{\infty}\frac{1}{x+1}\frac{e^{-\lambda}\lambda^x}{x!}. \end{align*}

I don't know how to proceed anymore so I went to see the given solution, the answer is $\frac{1-e^{-\lambda}}{\lambda}$.

Answer 1

Note

$$\frac{1}{x+1}\frac{e^{-\lambda}\lambda^x}{x!}=\lambda^{-1}e^{-\lambda}\color{red}{\frac{\lambda^{x+1}}{(x+1)!}}$$

Do you recognize the terms in red?