As observed in Find the expected value of $E[\frac{1}{\left(X+1\right)^2}]$ where X is binomial, there exists a closed form solution to $$\mathbb{E}\left[\frac{1}{(1+X)^2}\right]$$ where $X$ is binomial random variable i.e $X \sim \textrm{Bin}(n,p)$. I am interested in proving an upper bound which I expect to be $$\mathbb{E}\left[\frac{1}{(1+X)^2}\right] \leq \frac{c(p)}{n^2}$$ for $n>1$ where $c(p)$ is some constant dependent on $p$. Any ideas on how to approach this?
EDIT: After @NN2's answer, I changed the question from $O(1/n^2)$ to $c(p)/n^2$ for $n>1$.
As $X$ follows binomial distribution, we can write $X=\sum_{i=1}^n Z_i$ with $Z_i$ follows the Bernoulli distribution. Then, according to central limit theorem, we have
$$\frac{1}{n}X =\frac{1}{n}\sum_{i=1}^n Z_i \xrightarrow{n\rightarrow +\infty}N(p,p(1-p))$$
So,
\begin{align} \mathbb{E}\left[\frac{1}{(1+X)^2}\right] &= \mathbb{E}\left[\frac{1}{n^2}\frac{1}{(\frac{1}{n}+\frac{X}{n})^2}\right] \\ &= \frac{1}{n^2}\mathbb{E}\left[\frac{1}{(\frac{1}{n}+\frac{X}{n})^2}\right] \xrightarrow{n\rightarrow +\infty}\frac{1}{n^2}\mathbb{E}\left[\frac{1}{N^2(p,p(1-p))}\right]\\ \end{align}