I want to understand [this entry](https://mathoverflow.net/questions/145678/expected-value-of-logarithm-of-a-binomial-random-variable), but do not understand how the $\mathcal{O}\left(\frac{1}{n^2}\right)$ in the accepted answer comes into play.


I reproduce the question here: We have $x \sim \mathrm{Bin}\_{p,n}$ and want to approximate $E[\ln(x+1)]$ with a Taylor series. And I'm only interested in the case when the Taylor series is convergent for all values of $x$. (This condition implies $p> \frac{1}{2} $). Thus,
\begin{equation}
E[T\_{x\_0=np}\left(\ln(x+1)\right)]=\ln(np + 1) - \sum\_{i=1}\frac{(-1)^{i}}{i(np-1)^{i}}E[(x-np)^{i}]
\end{equation}
In the accepted solution, the suggestion is to use the following approximation:
\begin{equation}
E[\ln(x+1)]=\ln(np + 1) - \frac{np(1-p)}{2(np+1)^{2}} + \mathcal{O}\left(\frac{1}{n^2}\right)
\end{equation}
Where the second summand comes from the variance term $(i = 2)$. I have not found an upper bound for central moments of Binomial variates that could lead to the big Oh notation. Plugging in a closed form expression for the respective central moments from [this paper](https://www.researchgate.net/profile/Andreas-Knoblauch/publication/220222321_Closed-form_Expressions_for_the_Moments_of_the_Binomial_Probability_Distribution/links/5552fca708ae6943a86d8fb1/Closed-form-Expressions-for-the-Moments-of-the-Binomial-Probability-Distribution.pdf) does not lead me to this approximation.



Expected value of Taylor series with central moments of binomial variate
There are many ways to bound the Binomial central moments. A good (very general) estimate is provided by the Marcinkiweicz-Zygmund inequality, in the sharp form due to Burkholder [1], Writing $X-np$ as a sum of $n$ i.i.d. mean zero variables $Y\_i$ taking values
$1-p$ and $-p$, observe that the square function S(X) defined on page 87 in [1] is bounded by $n^{1/2}$. Theorem 3.1 page 87 in [1] then gives
$$E[|X-np|^k] \le \bigl((k-1)n^{1/2}\bigr)^k \,.$$  

The case $k=3$ is best dealt with directly, and here we cannot afford the absolute value inside the expectation:
$$E[(X-np)^3] =\sum\_{i,j,\ell \le n} E[Y\_iY\_jY\_\ell]=\sum\_{i \le n} E[Y\_i^3] \le n \,.$$


[1] Burkholder, Donald L. "Sharp inequalities for martingales and stochastic integrals." Astérisque 157, no. 158 (1988): 75-94.
<http://www.numdam.org/article/AST_1988__157-158__75_0.pdf>


Remark: In my answer I focus on the question regarding Binomial central moments. for estimating the logarithm, see Iosif Pinelis' answer.


