Fisher information of a Binomial distribution The Fisher information is defined as $\mathbb{E}\Bigg( \frac{d \log f(p,x)}{dp} \Bigg)^2$, where $f(p,x)={{n}\choose{x}} p^x (1-p)^{n-x}$ for a Binomial distribution. The derivative of the log-likelihood function is $L'(p,x) = \frac{x}{p} - \frac{n-x}{1-p}$. Now, to get the Fisher infomation we need to square it and take the expectation.
First, we know, that $\mathbb{E}X^2$ for $X \sim Bin(n,p)$ is $ n^2p^2 +np(1-p)$. Let's first focus on on the content of the paratheses.
$$
\begin{align}
\Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2&=\frac{x^2-2nxp+n^2p^2}{p^2(1-p)^2}
\end{align}
$$
No mistake so far (I hope!).
\begin{align}
\mathbb{E}\Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2 &= \sum_{x=0}^n \Bigg( \frac{x}{p} - \frac{n-x}{1-p} \Bigg)^2 {{n}\choose{x}} p^x (1-p)^{n-x} \\
&=\sum_{x=0}^n \Bigg( \frac{x^2-2nxp+n^2p^2}{p^2(1-p)^2} \Bigg) {{n}\choose{x}} p^x (1-p)^{n-x} \\
&= \frac{n^2p^2+np(1-p)-2n^2p^2+n^2p^2}{p^2(1-p)^2}\\
&=\frac{n}{p(1-p)}
\end{align}
The result should be $\frac{1}{p(1-p)} $ but I've been staring at this for a few hours incapable of getting a different answer. Please let me know whether I'm making any arithmetic mistakes.
 A: So, you have $X$ ~ Binomial($n$, $p$), with pmf $f(x)$:

You seek the Fisher Information on parameter $p$. Here is a quick check using mathStatica's FisherInformation function:

which is what you got :)
A: Fisher information: $I_n(p) = nI(p)$, and $I(p)=-\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg)$, where $f(p,x)={{1}\choose{x}} p^x (1-p)^{1-x}$ for a Binomial distribution. We start with $n=1$ as single trial to calculate $I(p)$, then get $I_n(p)$.
$\log f(p,x) = x \log p + (1-x) \log p$
$\frac {\partial \log f(p,X)}{\partial p} = \frac {X}{p} - \frac {1- X}{1 - p}$
$\frac {\partial^2 \log f(p,X)}{\partial p^2} = -\frac {X}{p^2} - \frac {1- X}{(1 - p)^2}$
$I(P) = -\mathbb{E_p}\Bigg( \frac{\partial^2 \log f(p,x)}{\partial p^2} \Bigg) = -\mathbb{E_p}\Bigg(-\frac {X}{p^2} - \frac {1- X}{(1 - p)^2}\Bigg) = \frac {p}{p^2} + \frac {1-p}{(1-p)^2} = \frac {1}{p} + \frac {1}{(1-p)} = \frac {1}{p(1-p)} $
As a result, $I_n(p) = n I(p) = \frac {n}{p(1-p)} $
A: You are correct.
In this case it is easier to find FI as -E d^2 {log f(x|p)}/dp^2.
A: The result does not dependent on $n$ in the asymptotic information matrix.
\begin{align*}
\mathcal{I}\left(p\right)&=\underset{n\to\infty}{\mathrm{plim}}\dfrac{1}{n}\dfrac{n}{p\left(1-p\right)}\\&=\dfrac{1}{p\left(1-p\right)}
\end{align*}
A: I  know this is well beyond time for the OP, but I have incurred into an analogous issue today and I would like to point out the source of confusion.
The Fisher information for a single Bernoulli trial is $\frac{1}{p(1-p)}$. When you have $n$ trial, the asymptotic variance indeed becomes $\frac{p(1-p)}{n}$. 
When you consider the Binomial resulting from the sum of the $n$ Bernoulli trials, you have the Fisher information that (as the OP shows) is $\frac{n}{p(1-p)}$. The point is that when you consider your variable as a Binomial you only have a sample of 1 -- since you observed only 1 binomial outcome. So that when you apply the classic result about the asymptotic distribution of the MLE, you have that the variance is simply the inverse of the Fisher information: $\frac{p(1-p)}{n}$ . 
Therefore, the asymptotic variances coincide from both perspectives.
