I'm trying to solve
$$ \sum_{b=0}^\infty Poisson(b, \lambda)\sum_{x=0}^b binomial(x, b, p)\left(\frac{x}{x+1}\right)^x \\ = \sum_{b=0}^\infty \frac{e^{-\lambda} \lambda^b}{b!}\sum_{x=0}^b {b \choose x}p^x(1-p)^{b-x}\left(\frac{x}{x+1}\right)^x$$
The trick of using the definition of the exponential doesn't work here as there is $b$ in the upper bound of the inner sum. If it weren't for that, I'd have some things I could try - but in this scenario, I am clueless.
Is there a closed form solution for this? How can I approach this problem?
EDIT: Let $S$ be your sum.
Interchange order of summations: $ S = \sum_{x=0}^\infty \sum_{b=x}^\infty \ldots $, and let $b=x+k$. You can then do the sum over $k$: the result is
$$ S = \sum_{x=0}^\infty \sum_{k=0}^\infty e^{-\lambda} \dfrac{\lambda^{x+k}}{x! k!} p^x (1-p)^k \left(\frac{x}{x+1}\right)^x = e^{-\lambda p} \sum_{x=0}^\infty \dfrac{1}{x!} \left(\dfrac{p\lambda x}{x+1 }\right)^x$$
This is unlikely to have a closed form.
BTW, for $x=0$ your summand involves $0^0$, which is not unambiguously defined (but that's not related to the lack of closed form).