I am trying to compute the following expectation, where $x$ follows a Poisson distribution, i.e.,: $$ \mathbb{E}[x] = \sum_{x=0}^{\infty} \frac{x}{x+a} \frac{\lambda^x e^{-\lambda}}{x!}, $$ where $\lambda > 0$, $a \in \mathbb{R}$ and $a > 0$.
I.e., I am wondering whether there is a closed-form expression for this summation. It clearly converges, since the fraction introduced in each term of the sum (weighted by the probability) is less than or equal to the value $x$ itself. However, I can't figure out how to reduce it. There are a lot of examples when $a$ is an integer, in which the denominator $x+a$ can be combined with the factorial in the denominator of the Poisson probability, but since $a$ may not be an integer, I don't see how it could be combined.
Similarly, I'm also trying to determine the following expectation as well:
$$ \mathbb{E}[x] = \sum_{x=0}^{\infty} \frac{a}{x+a} \frac{\lambda^x e^{-\lambda}}{x!}, $$ which has almost the same form. The $a$ in the numerator can be remove from the summation in the latter expectation, whereas the $x$ in the numerator in the former can be combined with the factorial, so I imagine if a solution exists for one the same sort of approach would allow a solution for the other.
I know that for deriving mean and such, usually the approach is to reduce the infinite sum to just include $\lambda^x / x!$ so it can be replaced by $e^\lambda$, but I can't see how to do it here.
The normalized Poisson distribtion is $$f(k)=\frac{t^k}{k!} e^{-t}, t>0$$ Then the expectation value of $1/(k+a)$ is given by $$E(1/(k+a))=\sum_{k=0}^{\infty} \frac{1}{k+a} \frac{t^k}{k!} e^{-t}=\sum_{k=0}^{\infty} \int_{0}^{1} u^{k+a-1}\frac{t^k}{k!} e^{-t} du=e^{-t}\int_{0}^{1} \sum_{k=0}^{\infty} \frac{(tu)^k}{k!} u^{a-1} du$$ $$\implies E(1/(k+a))= e^{-t} \int_{0}^{1} e^{tu} u^{a-1} du=e^{-t}[ (-t)^{-a}~ [\Gamma(a)-\Gamma(a,-t)]$$ Next $$E(a/(k+a))= ae^{-t} (-t)^{-a}~ [\Gamma(a)-\Gamma(a,-t)]$$ And $$E(k/(k+a))=1- ae^{-t} (-t)^{-a}~[\Gamma(a)-\Gamma(a,-t)]$$ Finally, $$ (-t)^{-a}[\Gamma(a)-\Gamma(a,-t)] = \sum_{k=0}^{\infty} \frac{t^k}{k!(k+a)},$$ which is all real for $t>0$.