Let $X\sim\pi(\lambda)$, find $E\Bigl[\frac{1}{(X+1)}\Bigr]$.
It is a problem in my text book, here is my work: $$\begin{align} E\left[\frac{1}{(X+1)}\right]&=\sum_{x=0}^{\infty}\frac{1}{x+1}\cdot\frac{\lambda^xe^{-\lambda}}{x!}\\ &=\frac{1}{\lambda}\sum_{x=0}^{\infty}\frac{\lambda^{x+1}e^{-\lambda}}{(x+1)!}\\ &=\frac{1}{\lambda}\sum_{k=1}^{\infty}\frac{\lambda^ke^{-\lambda}}{k!}\space (let \space k=x+1)\\ &=\frac{1}{\lambda}\left[\sum_{k=0}^{\infty}\frac{\lambda^ke^{-\lambda}}{k!}-e^{-\lambda}\right]\\ &=\frac{1}{\lambda}\left[1-e^{-\lambda}\right]\\ \end{align}$$
Is it right?
Although you have solved it, here's another way to solve this question. Use MGF $E(e^{tx})=e^{\lambda(e^{t}-1)}$
Similarly $E(s^x)=e^{\lambda(s-1)}$
$\dfrac{1}{X+1}=\int_{0}^{1} s^Xds$
$E\left(\dfrac{1}{X+1} \right)=\int_{0}^{1} E(s^X)ds=\int_{0}^{1} e^{\lambda(s-1)}ds=\bigg(\dfrac{e^{\lambda s}e^{-\lambda}}{{\lambda}}\bigg)_{0}^{1}=\dfrac{e^{-\lambda}(e^{ \lambda }-1)}{{\lambda}}=\dfrac{1}{\lambda}\left[1-e^{-\lambda}\right]$