# Operator product expansion of more than 2 operators in CFT

I’m confused about OPE in 2d CFT. I’ve found difficulties in taking OPE of the product of 3 operators.

Consider the following operator product.

\begin{align*} O_1(z) :O_2(w) O_3(w): = O_1(z)\frac{1}{2\pi i}\oint_{w} \frac{dx}{x-w}O_2(x)O_3(w) \end{align*} Where I used contour integral to define the holomorphic part of $O_2(w)O_3(w)$

You can expand the $O_1(z)O_2(x)$ product first.Let me assume their OPE, \begin{align*} O_1(z)O_2(x)= \sum_{n < 0} (z-x)^n O_{12}^{n}(x) + \sum_{n \geq 0} (z-x)^n O_{12}^{n}(x) \end{align*} Where divergent terms and finite terms are separated.

The first divergent terms would contribute as the contraction of $O_1(z)O_2(w)$ in $O_1(z):O_2(w)O_3(w):$. So I thought the second terms eventually give the contribution of the contraction of $O_1(z)O_3(w)$ after calculating the integral, $$\frac{1}{2\pi i}\oint_{w}\frac{dx}{x-w} \sum_{n \geq 0} (z-x)^n O_{12}^{n}(x)O_3(w)$$

But this does not become the divergent term because n runs from 0 and contour integral remove the divergence from the contraction of $O_{12}^n(x)O_3(w)$.

According to the textbook I’ m reading, this seems to be wrong because the method I explained does not give the correct answer. So what is wrong with this argument?

Is it too naive to think you can get the full OPE by taking two operator OPE one after the other from the left? For example, is there any rule about the order of taking OPE depending on the arguments of operators?