Expected value of logarithm of a binomial random variable Suppose $X\sim \mathrm{Binomial}(n,p),$ i.e. $X$ takes values in $\{0,1,2,\ldots, n\}$ and $P(X=i) = {n\choose i} p^i(1-p)^{n-i}.$
I am looking for a good estimate for $\mathbb{E}\log(X+\alpha).$ I am thinking of $\alpha$ as some fixed number, for example, $\alpha = 1,$ $p$ as a fixed number between 0 and 1, eg. $p=\frac{1}{3}$ while $n$ is some very large number. The kind of estimate I need may be clarified by the following aim:
Define $f(n): = \log(n+2\alpha) +p\mathbb{E}\log\frac{p}{X+\alpha} + (1-p)\mathbb{E}\log\frac{1-p}{n-X+\alpha}.$ I want to show that $f(n)\sim \frac{c}{n} + o\left(\frac{1}{n}\right)$ and I want to identify the constant $c.$
 A: Just expand $ln(X+\alpha)$ as a Taylor series about $X=np$ and then do the binomial sum term by term. Use tail bounds on the binomial distribution to show that the error terms are meaningful. Without much checking, I got
$$\mathbb E(\ln(X+\alpha)) = \ln(pn+\alpha) - \frac{1-p}{2pn} + O(1/n^2),$$
which ought to be precise enough for you.  More terms are easy to get.
From there your answer will follow on expanding and approximating some logs.
ADDED: In response to Hedonist's comments, here is a tiny bit more detail.  We know from the Chernov inequality, or otherwise, that $\mathrm{Prob}(|X-np|\gt n^{1/2+\epsilon})\lt e^{-cn^{2\epsilon}}$ for small enough $\epsilon\gt 0$ and some $c\gt 0$.  (I'm assuming $p$ is constant, or at least bounded away from 0 and 1.) Since $\ln(X+\alpha)=O(\ln n)$ for all $X$, this means the contribution of the values of $X$ outside that range is at most $O(\ln n)e^{-cn^{2\epsilon}}$, which is negligible. Within the interval $|X-np|\le n^{1/2+\epsilon}$ you can expand the logarithm by Taylor series and easily see how many terms are needed for the precision you want.
