Approximate expectation of a random variable that is the logarithm of a function of a binomial I want the expectation of the following random variable: $\log\left(\frac{X}{k-X}+\alpha \right)$ with $X \sim Bin_{(k-1),p}$ and $\alpha > 0$, Therefore I derived the Taylor Series:
\begin{equation}
T_{x_0=(k-1)p}\left[\log\left(\frac{X+\alpha}{k-X}\right)\right] = \log\left(\frac{x_0+\alpha}{k-x_0}\right)+ \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(-1)^{n-1}}{(x_0+\alpha)^n}+\frac{1}{(k-x_0)^n}\right)\left(x-x_0\right)^n 
\end{equation}
Plugging in $x_0=(k-1)p$ and calculating the expectation leads me to:
\begin{equation}
E\left[\log\left(\frac{X+\alpha}{k-X}\right)\right]= \log\left(\frac{(k-1)p+\alpha}{k-(k-1)p}\right)+\sum_{n=1}^{\infty}\frac{1}{nk^n}\left(\frac{(-1)^{n-1}}{\left(p+\frac{\alpha - p}{k}\right)^n}+\frac{1}{\left(1-p+\frac{p}{k}\right)^n}\right)E\left[\left(x-(k-1)p\right)^n\right]
\end{equation}
I know that the Taylor series of $\log(x+1)$ only converges within the open ball $(-1,1)$. Does that apply for the random variable? That is, that for $f(X) = \log\left(\frac{X+\alpha}{k-X}\right)$ it holds that:
\begin{equation}
E\left[f(X)\right]=E\left[T_{x_0=(k-1)p}f(X)\right] \Leftrightarrow f(X) \in (-1,1)
\end{equation}
Clearly, that would depend on the values of $k$ and $p$. However, for me $k$ can take values up to a few hundred, so $f(X) \in (-1,1)$ does not hold $\forall X$.
When I wasn't aware that this convergence could go wrong, I showed that in the parameter settings $p=0.5$ and $k>>1$, so that $k \approx (k-1)$, the expectation is $0$, given the Taylor approximation is true. Now, can I still use this result from the Taylor series?
 A: So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[\log\left(X+\alpha\right)\right]-E\left[\log\left(Y+1\right)\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test.
\begin{equation}
T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i
\end{equation}
My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is:
\begin{equation}
r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha
\end{equation}
Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p\geq\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.
So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that
\begin{equation}
\lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0    
\end{equation}
as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.
