Answer is (C)
\begin{align*}
E\left( \frac{1}{1+x}\right ) &= \sum_{k = 0}^{\infty} \left( \frac{1}{1+k}\right ) * \frac{\lambda^k*e^{-\lambda}}{k!} \\
&= \frac{1}{\lambda} * \sum_{k = 0}^{\infty} \frac{\lambda^{k+1}*e^{-\lambda}}{(k+1)!}\\
&= \frac{e^{-\lambda}}{\lambda} * \sum_{k = 1}^{\infty} \frac{\lambda^{k}}{k!}\\
&= \frac{e^{-\lambda}}{\lambda} * \left( \sum_{k = 0}^{\infty} \frac{\lambda^{k}}{k!}-1\right)\\
&= \frac{e^{-\lambda}}{\lambda} * \left( e^{\lambda} -1\right)\\
&= \frac{1-e^{-\lambda}}{\lambda}
\end{align*}