Expectation & Variance of $\frac{1}{K+X}$ when $X \sim $ Poisson and $K=2,3,4 \ldots$ I recently posted a question about how to find the variance of $\frac{1}{1+X}$ where X has a Poisson distribution. Variance of $\frac{1}{X+1}$ when $X$ has Poisson distribution 
And got a very helpful answer. 
(others had already answered how to get the expectation of same Expectation of $\frac{1}{x+1}$ of Poisson distribution ; Expected value $\frac{1}{x+1}$ of Poisson distribution ) 
I'm now realizing that I also need the expectation and variance of $\frac{1}{K+X}$ where X is Poisson, for integer values of $K>1$ too. Is there a general form of the solution for getting the expectation and variance that allows me to do this for these larger integer values of K? (In practice, I'm only going to need at most up to $K=5$ but I suspect solving for positive integer $K$ in general will be easier than solving for $K=2,3,4,5$ separately.)
I tried extending the answers for $K=1$ to $K=2$ and ran into a wall. Any help is greatly appreciated. 
 A: Let $X\sim\mathrm{Pois}(\lambda)$. We compute the mean of $\frac1{K+X}$:
\begin{align}
\mathbb E\left[\frac1{K+X}\right] &= \sum_{j=0}^\infty \frac1{K+j} \mathbb P(X=j)\\
&= \sum_{j=0}^\infty \frac1{K+j} e^{-\lambda}\frac{\lambda^j}{j!}\\
&= e^{-\lambda } (-\lambda )^{-K} \Gamma (K,0,-\lambda ),
\end{align}
where $\Gamma(K,0,-\lambda) = \int_0^{-\lambda} t^K e^{-t}\ \mathsf dt $ is the generalized incomplete gamma function. The second moment is given by
\begin{align}
\mathbb E\left[\left(\frac1{K+X}\right)^2\right] &= \sum_{j=0}^\infty \left(\frac1{K+j}\right)^2 e^{-\lambda}\frac{\lambda^j}{j!}\\
&= \frac{e^{-\lambda } \, _2F_2(K,K;K+1,K+1;\lambda )}{K^2},
\end{align}
where ${}_2F_2(K,K;K+1,K+1;\lambda)$ is the generalized hypergeometric function.
The variance is given by
\begin{align}
\mathbb E\left[\left(\frac1{K+X}\right)^2\right] - \mathbb E\left[\frac1{K+X}\right]^2 &= \frac{e^{-\lambda } \, _2F_2(K,K;K+1,K+1;\lambda )}{K^2} - (e^{-\lambda } (-\lambda )^{-K} \Gamma (K,0,-\lambda ))^2\\
&= e^{-2 \lambda } \left(\frac{e^{\lambda } \, _2F_2(K,K;K+1,K+1;\lambda )}{K^2}-(-\lambda )^{-2 K} \Gamma (K,0,-\lambda )^2\right).
\end{align}
