\mathrm{E}\left(\frac{1}{1+X}\right)=\sum\limits_{n=0}^{\infty}{\frac{1}{1+n}\cdot\mathbb{P}{\left(X=n\right)}}=\sum\limits_{n=0}^{\infty}{\frac{1}{1+n}\cdot \frac{\lambda^n}{n!}e^{-\lambda}}=