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? 
 A: A sum of $K$ unit fractions, each of denominator $n_i \leq n$, can 
be rewritten as a fraction with a denominator bounded by 
the product of the $n_i$, i.e. by $n^K$. (A small improvement is possible, with product of distinct integers $\leq n$)
A difference of two such fractions
(which are themselves sums of $K_1$ and $K_2$ fractions, respectively),
is a fraction with a demominator bounded by $n^{K_1+K_2}$.
(If $K_1\neq K_2$ the order of magnitude of the difference is actually
$\frac{1}{n^{\min(K_1,K_2)}}$.)
Let us assume that $K_1=K_2$.
So, the smallest nonzero difference $d(K,n)$ of sums of 
$K$  unit fractions, 
with $n_i \leq n$ is $d(K, n)\leq \frac{1}{n^{2K}}$.
I believe that this order of magnitude, with slightly weaker constants,
 can be achieved with a constructive parametrization.
Example: If $K_1=K_2=2$, then choose
$\frac{1}{x^2 + 4 x + 1} + \frac{1}{x^2 + 4 x + 3} - 
\frac{1}{x^2 + 3 x + 1} - \frac{1}{x^2 + 5 x + 5}=
\frac{2}{(1 + 3 x + x^2) (1 + 4 x + x^2) (3 + 4 x + x^2) (5 + 5 x + x^2)}$.
Now, taking $n=x^2+5x+5$, one has a set of 4 unit fractions,
for which the difference of the sums above is asymptotically
$ \frac{2}{n^4}$. This example is (for $K_1=K_2=2$)
possibly the best one can find, (but I did not prove this).
I conjecture that for larger values of 
$K$ one can construct similar polynomial examples.
Quite possibly this has applications in questions in diophantine approximation,
exponential sums, large sieve etc. 
A: 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
$\frac{1}{10x} + \frac{1}{x-2} + \frac{1}{2x-8} - \frac{1}{2x-2} - \frac{1}{x-3} - \frac{1}{10x-50} = \frac{-12}{x(x-1)(x-2)(x-3)(x-4)(x-5)}$.
A: if $k_1=k_2$
$A=\biggl|(\sum_{i=1}^{K_1} \frac{1}{a_i})-(\sum_{j=1}^{K_1} \frac{1}{b_j})\biggr|\le\sum_{i=1,j=1}^{K_1}\biggl|\frac{1}{a_i}-\frac{1}{b_j}\biggr|$
we assume that $a_i=n$ and $b_j=n-1$,so $A\le \frac{k_1}{n(n-1)}$
if $k_1\le k_2$,and assume that$a_i=n$ and $b_j=n-1$,so $A\le \frac{k_2}{n(n-1)}$
Added:let $a_i=b_i+b_i^{mk-2}$,$mk\ge 2$,$m$ is positive integer ,$1\le b_i\le n$, and $k_1\ge k_2$,so 
$\biggl|(\sum_{i=1}^{K_1} \frac{1}{b_i})-(\sum_{i=1}^{K_2} \frac{1}{a_i})\biggr|$,this is
larger or equal to ,$\sum_{i=1}^{K_1}\frac{b_i^{mk-2}}{b_i(b_i+b_i^{mk-2})}\ge\sum_{i=1}^{K_1}
\frac{1}{b_i(b_i+b_i^{mk-2})}\ge\frac{k_1}{n^{mk}}$
A: See http://en.wikipedia.org/wiki/Egyptian_fraction for some background. You are asking about the difference of rational numbers which can be any positive rationals. So the lower bound is in terms of what? It is natural to ask in terms of the maximum size of required denominator. I think the bound in that article in "Modern number theory" probably answers that question. It should come down to how far apart you can be sure the Farey sequence fractions are (which is known and elementary): http://en.wikipedia.org/wiki/Farey_sequence .
