Help with Taylor expansion of log likelihood function I've reed the following part of a sketch of the proof that the maximum likelihood estimator is asymptotically normal:
"Sketch of the second part of the proof.
Recall that we may write the likelihood equation as
$$\sum_{i=0}^n U_i(\hat\theta)=0$$  where $  U_i(\phi)$ denotes the derivative of $ \log\! f(Y_i;\phi)  $ with respect to $\phi$.
Let $ U_i'(\phi) $ denote the derivative of     $ \log\!f(Y_i;\phi)$ with respect to $ \phi$ . Now a Taylor expansion around $ \phi=\theta$ yields:
$$\sum_{i=0}^n  U_i(\phi )-\sum_{i=0}^n  U_i(\theta)\approx \left( \sum_{i=0}^n U_i'(\theta) \right)(\phi-\theta)$$"
This Taylor expansion does not make any sense to me. I am familiar with a Taylor expansion of $f(x) $ at $a$ as: $$\sum_{i=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$
I can see that all terms except the second term from the Taylor expansion of the log likelihood are excluded but I do not recognise where this term comes from as the original term involved subtracting two summations.
 A: You should convince yourself that $f(x)-f(y)\approx f'(x)(x-y)$ is just another way to express Taylor expansion (under appropriate regularity assumptions).
Then, using linearity of the derivation, you can generalize it to the sum.
A: I think your problem is just the notation problem
Let us start from the beginning;
Suppose$X_1, X_2,...,X_n$ are i.i.d random variables with probability density function $f(x;\gamma)$
Usually, people will use $\theta$ here, I will keep it for later use.
The likelihood function is $L(\gamma;x)=f(x_1;\gamma)f(x_2;\gamma)...f(x_n;\gamma)$
The log likelihood function $l(\gamma;x)=log[f(x_1;\gamma)f(x_2;\gamma)...f(x_n;\gamma)]=\sum_{i=1}^nlogf(x_i;\gamma)$
As usual, we will take derivative of the log likelihood and   set it to zero i.e $l'(\gamma;x)=0 $
or $\sum_{i=1}^nlog'f(x_i;\gamma)=0$
Here, you can see  $U_i(\gamma)=log'f(x_i;\gamma)$
$\therefore l'(\gamma)=\sum_{i=1}^nU_i(\gamma)\tag{1}$ we omit $x$ here since the log likelihood function is a function of $\gamma $ 
Next we expand the function $l'(\gamma)$ into a Taylor series of order two abut $\theta$ .
$l'(\gamma)=l'(\theta)+\frac{l''(\theta)}{1!}(\gamma-\theta)^1+\frac{l'''(\theta)}{2!}(\gamma-\theta)^2$. This is the Taylor expanding for $l'(\gamma)$.
Next we evaluate the equation at $\phi$
$l'(\phi)=l'(\theta)+\frac{l''(\theta)}{1!}(\phi-\theta)^1+\frac{l'''(\theta)}{2!}(\phi-\theta)^2 \tag{2}$
Here you should see that $l'(\theta)=\sum_{i=1}^nU_i(\theta)$ 
and $l'(\phi)=\sum_{i=1}^nU_i(\phi)$
and $l''(\phi)=\sum_{i=1}^nU_i'(\phi)$
Ref (1)
If we ignore the third derivative term in (2) then your question will be answered here
We get that $\sum_{i=1}^n  U_i(\phi )-\sum_{i=1}^n  U_i(\theta)\approx \left( \sum_{i=1}^n U_i'(\theta) \right)(\phi-\theta)$
By the way I think $i$ usually start from $1$ not $0$ ,anyway it is just an index.
Let us don't stop here, we can go further to prove the theorem
We know that $l'(\phi)=0$
$\therefore l'(\theta)+\frac{l''(\theta)}{1!}(\phi-\theta)^1+\frac{l'''(\theta)}{2!}(\phi-\theta)^2=0$
i.e.
$l'(\theta)+(\phi-\theta)*[l''(\theta)+\frac{l'''(\theta)}{2}(\phi-\theta)]=0$
Next we rearrange the above terms:
$$(\phi-\theta)=\frac{l'(\theta)}{-l''(\theta)-\frac{l'''(\theta)}{2}(\phi-\theta)}$$
We multiply $\sqrt{n}$ for both side:
$$\sqrt{n}(\phi-\theta)=\frac{\sqrt{n}*l'(\theta)}{-l''(\theta)-\frac{l'''(\theta)}{2}(\phi-\theta)}\\=\frac{\frac{1}{\sqrt{n}}*l'(\theta)}{\frac{-l''(\theta)}{n}-\frac{l'''(\theta)}{2n}(\phi-\theta)} \tag{3}$$ 
(divide by n for numerator and denominator at the same time for the left hand side)
The let us see what is the numerator of the left side of (3):
$$\frac{1}{\sqrt{n}}l'(\theta)=\frac{1}{\sqrt{n}}\sum_{i=1}^n\frac{\partial logf(x_i;\theta)}{\partial \theta}$$
And note $$\frac{\partial logf(x_i;\theta)}{\partial \theta}$$ are i.i.d with variance $I(\theta)$ and $$E(\frac{\partial logf(x_i;\theta)}{\partial \theta})=0$$
$\therefore$ by CLT
$$\frac{1}{\sqrt{n}}l'(\theta)\sim \frac{1}{\sqrt{n}}N(0,nI(\theta))=N(0,I(\theta))$$
Next we will see what are in the denominator of (3):
$$-\frac{l''(\theta)}{n}=-\frac{1}{n}\sum_{i=1}^n\frac{\partial^2 logf(x_i;\theta)}{\partial \theta}\overset{P}{\rightarrow} I(\theta)$$ by Law of Large number.
For the term $$\frac{l'''(\theta)}{2n}(\phi-\theta)$$ in denominator of (3) we can prove it convergence in Probability to zero. 
Finally, let us warp up everything
$$\sqrt{n}(\phi-\theta)\sim \frac{N(0,I(\theta)}{I(\theta)}=N(0,\frac{1}{I(\theta)})$$
This proved the theorem.
