<s>[INST] The difference of two sums of unit fractions I had this question bothering me for a while, but I can't come up with a meaningful answer. 
The problem is the following:
Let integers $a_i,b_j\in${$1,\ldots,n$} and $K_1,K_2\in$ {$1,\ldots,K$}, then how small (as a function of $K$ and $n$), but strictly positive, can the following absolute difference be.
$\biggl|(\sum_{i=1}^{K_1} \frac{1}{a_i})-(\sum_{j=1}^{K_2} \frac{1}{b_j})\biggr|$
As an example for $K_1=1,$$K_2=1$ choosing $a_1=n,$$b_1=n-1$gives the smallest positive absolute difference, that is $\frac{1}{n(n-1)}$. What could the general case be? 
 [/INST] We can prove Christian Elsholtz's conjecture using only linear functions.
Let $m_1, ..., m_{2K}$ and $\alpha_1, ..., \alpha_{2K}$ be integers satisfying $\sum_i \frac{\alpha_i^j}{m_i} = 0$ for $j = 0, ..., 2K-2$. Then for large $x$ we have
$\sum_i \frac{1}{m_ix - m_i\alpha_i} = \frac{1}{x}\sum_i\frac{1}{m_i}\sum_j\frac{\alpha_i^j}{x^j} = \sum_j \frac{1}{x^{j+1}} \sum_i \frac{\alpha_i^j}{m_i} \approx \frac{C}{x^{2K}}$,
where $C = \sum_i \frac{a_i^{2K-1}}{m_i}$.
For instance, we can pick $\alpha_i = i-1, m_i = \frac{(-1)^iD}{\binom{2K-1}{i}}$, where $D$ is a multiple of $\binom{2K-1}{i}$ for each $i$. Then we will have the same number of positive and negative $m_i$s, so for large $x$ half of the fractions will be positive and half will be negative.
Examples:
$\frac{1}{3x} + \frac{1}{x-2} - \frac{1}{x-1} - \frac{1}{3x-9} = \frac{-2}{x(x-1)(x-2)(x-3)}$, and
$\frac{1}{10x} + \frac{1}{x-2} + \frac{1}{2x-8} - \frac{1}{2x-2} - \frac{1}{x-3} - \frac{1}{10x-50} = \frac{-12}{x(x-1)(x-2)(x-3)(x-4)(x-5)}$.
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