# Why do we exclude the $(i,i)$ case when summing over internal forces?

In the majority of the literature and lectures I see when a system of particles is involved, I usually see the following expression (or similar) for the total force on particle $i$:

$$\vec{F}_i = \vec{F}_{i_{ext}} + \sum_{i\neq j}\vec{F}_{ji}$$

Why is it necessary to include the condition $i\neq j$ in the internal force terms? We know that $\vec{F}_{ii} = 0$, and hence removing the constraint and summing over general $i,j$ surely won't change the total.

If $F_{ii}=0$ then you are right that the $i\neq j$ constraint is unnecessary although it does make the physical interpretation clearer: all atoms (other than $i$) act on $i$.
However, in practice, the force is usually given as a function of separation, $F(r)$. And so when you evaluate $F_{ii}$, you effectively evaluate $F(0)$. The problem is that $F(0)$ is almost never zero.
For instance, the Coulomb and Lennard-Jones interactions are undefined for $r=0$ and so, when evaluating them (either by hand or in code), you must explicitly skip the $i=j$ case.