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Sum of reciprocals

Let x_1<x_2<\cdots<x_n be points of [-1, 1]. Prove that

    \[\sum_{1\leq j<k\leq n}\frac{1}{x_k-x_j} = \frac{1}{2} n^2\log n + \mathcal{O} \left ( n^2 \right )\]

Solution

First of all we note that

    \[\sum_{1 \leq j < k\leq n} \frac{1}{x_k - x_j} = \sum_{r=1}^n \sum_{s=1}^{n-r} \frac{1}{x_{s+r} - x_s}\]

It now follows from the arithmetic – harmonic inequality that

    \begin{align*} \sum_{s=1}^{n-r} \frac{1}{x_{s+r} - x_s} &\geq \frac{(n-r)^2}{\sum_{s=1}^{n-r} (x_{s+r} - x_s)} \\ &= \frac{(n-r)^2}{\sum_{t=1}^{r}(x_{n+1-t}-x_t)} \\ & \geq \frac{(n-r)^2}{2r} \end{align*}

And finally,

    \begin{align*} \sum_{1 \leq j < k\leq n} \frac{1}{x_k - x_j} &\geq \frac{1}{2}\sum_{r=1}^n \left(\frac{n^2}{r} - 2n + r \right) \\ &= \frac{1}{2}n^2 \log{n} + \mathcal{O} \left(n^2 \right) \end{align*}

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