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For example, $$ \begin{align} S_1&=\frac{x_1}{2},\\ S_2&=\frac{x_1}{2}-\frac{x_1^2}{6}\frac{1}{x_2},\\ S_3&=\frac{x_1}{2}-\frac{x_1^2}{6}\left(\frac{1}{x_2}+\frac{1}{x_3}\right)+\frac{x_1^3}{12}\frac{1}{x_2 x_3},\\ S_4&=\frac{x_1}{2}-\frac{x_1^2}{6}\left(\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}\right)+\frac{x_1^3}{12}\left({\frac{1}{x_2 x_3}+\frac{1}{x_2 x_4}+\frac{1}{x_3 x_4}}\right)-\frac{x_1^4}{20}\frac{1}{x_2 x_3 x_4},\\ &\cdots \end{align} $$I am trying to generalize this using $\sum$ and $\prod$ notation. My attempt: $$S_n=\sum_{i=1}^{n}{\frac{(-1)^{i-1}x_1^i}{i(i+1)}\sum_{2\le j_1\lt j_2\lt\cdots\lt j_{i-1}\le n}\prod_{k=1}^{i-1}\frac{1}{x_{j_k}}},$$ which still seems complicated. I wonder if there are more elegant expressions. :-)

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    $\begingroup$ Please don't silently change the question after it has been answered. $\endgroup$ – Yves Daoust Jun 21 at 10:29
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    $\begingroup$ @YvesDaoust Sorry for the confusion! There was no answer yet when I started editing the question. I just added the expressions for $S_1,S_2,S_3$ to describe my question more clearly. Please accept my apology. $\endgroup$ – MinutesSneezer Jun 21 at 10:44
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One way is the express it as $$ S_n := \frac1{x_1}\int_0^{x_1} \int_0^{t} \prod_{k=2}^n \left(1 - \frac{x}{x_k}\right) \, \text{d}x \, \text{d}t. $$ The clue is the appearance of elementary symmetric polynomials in the coefficients of $\,x_1^k$.

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  • $\begingroup$ For clarification, I suggest writing it as $$S_n=\frac1{x_1}\int_0^{x_1}\int_0^t\prod_{k=2}^n\left(1-\frac{x}{x_k}\right)\,\mathrm{d}x\,\mathrm{d}t.$$ I wonder how you managed to observed this. Is it related to polynomials? $\endgroup$ – MinutesSneezer Jun 21 at 11:16
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    $\begingroup$ @MinutesSneezer Thanks for those improvements! The finite product is a polynomial in $x$ if that is what you mean. $\endgroup$ – Somos Jun 21 at 12:00
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Probably not what you are after, but observing that $\dfrac1{x_k}, k=2,\cdots4$ are the roots of some polynomial

$$\left(z-\frac1{x_2}\right)\left(z-\frac1{x_3}\right)\left(z-\frac1{x_4}\right)=p_3z^3+p_2z^2+p_1z+p_0,$$

you have

$$S_4=\sum_{i=1}^4\frac{p_{4-i}}{i(i+1)}x_1^i.$$


You can also refer to the elementary symmetric polynomials notation.

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  • $\begingroup$ Exactly needed! Using the notation described in Wikipedia, it simply becomes $$\sum_{i=1}^{n}{\frac{(-1)^{i-1}x_1^i}{i(i+1)}e_{i-1}\left(\frac{1}{x_2},\frac{1}{x_3},\cdots,\frac{1}{x_n}\right)}.$$ $\endgroup$ – MinutesSneezer Jun 21 at 10:43

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