Reciprocal of a Binomial I'm wondering if there is any formula to do this.
Suppose $B$~$B(N,p)$, and hence we have $E[B]=Np,D[B]=Np(1-p)$. I'm just wondering how to do $E[\frac{1}{c+B}]$,for some $c>0$?
Thanks. 
 A: Law of the unconscious statistician: 
$$
\mathbb E[g(X)] = \sum_xg(x)\mathbb P(X=x)
$$
In this case, we have the sum:
\begin{align}
\mathbb E\left[\frac1{c+B}\right]&=\sum_{n=0}^N\frac1{c+n}\mathbb P(B=n)\\
&=\sum_{n=0}^N\begin{pmatrix}N\\n\end{pmatrix}\frac{p^n(1-p)^{N-n}}{c+n}
\end{align}
According to Wolfram|Alpha, this is equal to: 
$$
\frac{(1-p)^N\,_2F_1\left(c, -N; c+1; \frac{p}{p-1}\right)}c
$$
where $\,_2F_1$ is the Hypergeometric function.  Since pretty much any function you've ever heard of is a special case of the hypergeometric function, this isn't a particularly interesting result, and I doubt there's any nice way to find it.  
A: A closed-form expression and the answer to this question is given in the paper by Chao, M. T., and W. E. Strawderman. “Negative Moments of Positive Random Variables.” Journal of the American Statistical Association, vol. 67, no. 338, 1972, pp. 429–431. JSTOR, www.jstor.org/stable/2284399.
A: For large $n$ (switching form $N$ to $n$), the following bounds might be useful.
Let us assume $p \in (0, 1/2]$. Then, for $B \sim \text{Bin}(n,p)$ and $\delta \in (0,p)$, we have
$$
\mathbb P( B \le \delta n) \le e^{-n D(\delta||p)}
$$
where $D(\delta||p)$ is the KL divergence between probability vectors $(\delta, 1-\delta)$ and $(p,1-p)$.
It follows that for all $n \ge 1$ and $0 < \delta < p \le 1/2$,
$$
\frac1{c+np} \;\le\; \mathbb E \Bigl[ \frac1{c+ B}\Bigr] \;\le \;
\frac{1}{c + n \delta} (1-e^{-n D(\delta||p)}) + \frac1{c} e^{-n  D(\delta||p)}
$$
where the lower bound is by the Jensen's inequality and the upper bound is by conditioning on the event $\{B > \delta n\}$ and its complement and bounding $1/(c+B)$ on the two events by $1/(c+n\delta)$ and $1/c$, respectively.

Another useful upper bound:
$$
\frac1{1+np} \;\le\; \mathbb E \Bigl[ \frac1{1+ B}\Bigr] \;\le \;
\frac{1}{(1+n)p}.
$$
