$$\frac{\sum_{i=1}^{n}x_{i}}{\prod_{i=1}^{n}x_{i}}= \frac{1}{x_{2}x_{3}x_{4}...}+\frac{1}{x_{1}x_{3}x_{4}}+\frac{1}{x_{1}x_{2}x_{4}}...$$
There is a very clear pattern that each consecutive result will be missing the next product in its denominator. Is there a simpler closed form representation of this result? Maybe a combinatorics formula with factorials or harmonic numbers?
If you are having trouble with understanding this, look more closely at the location of the $i$ index to notice it is in the sub-script, not the super-script. If such an index was in the super-script, and you assumed $x$ was a real variable, then and only then could each term be assumed as a mononomial resulting in an overall polynomial in the sum, followed by reciprocals of mononomials on the right-hand side.
Assuming the $x_i$ are invertible elements of some ring $R$, their sum divided by their product is the $(n-1)$-th elementary symmetric polynomial in the $x_i^{-1}$, as noted in the comments. That is to say, $$\frac{\sum_{i=1}^{n}x_{i}}{\prod_{i=1}^{n}x_{i}}=\sum_{\sigma\in S_n}\prod_{i=1}^{n-1}x_{\sigma(i)}^{-1}=:e_{n-1}(x_1^{-1},\ldots,x_n^{-1}),$$ where the latter is simply a commonly used notation for elementary symmetric polynomials. It is worth noting that all elementary symmetric polynomials are irreducible, so this expression does not factor in any way.
This also shows that it is the $X$-coefficient of the polynomial $$f:=\prod_{i=1}^n(X-x_i^{-1})\in R[X],$$ or equivalently that it equals $f'(0)$.