$$\frac{\sum_{i=1}^{n}x_{i}}{\prod_{i=1}^{n}x_{i}}= \frac{1}{x_{2}x_{3}x_{4}...}+\frac{1}{x_{1}x_{3}x_{4}}+\frac{1}{x_{1}x_{2}x_{4}}...$$

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.

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    $\begingroup$ Without knowing anything specific about the $x_i$'s? No. $\endgroup$ – darij grinberg Feb 7 at 5:24
  • $\begingroup$ The only important thing about the $x_{i}$s is they cancel out no matter what number or variable they are, except zero. So, "no, as far as I know" is a more appropriate answer, which would have been assumed anyway had you not commented. This could also be related to the harmonic series. $\endgroup$ – Vane Voe Feb 7 at 5:26
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    $\begingroup$ What do you mean by "cancel out"? Your fraction is an irreducible polynomial in the $x_i^{-1}$ as soon as $n \geq 2$; indeed, it is the $\left(n-1\right)$-st elementary symmetric polynomial in them. Any "formula" for this fraction would translate into a formula for the $\left(n-1\right)$-st elementary symmetric polynomial in $n$ arbitrary variables. $\endgroup$ – darij grinberg Feb 7 at 5:28
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    $\begingroup$ @darijgrinberg I think that is quite a satisfactory answer. $\endgroup$ – Servaes Feb 7 at 5:30
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    $\begingroup$ Dear Vane Voe and @Servaes, please keep your interactions civil. Do not retaliate if you find yourselves in such situation again, use the flag system accordingly. I deleted comments that were out of place. Regards, $\endgroup$ – Pedro Tamaroff Feb 8 at 21:12

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 $f'(0)$.


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