I am trying to prove that $\sum_{i=1}^{n}\frac{1}{a_i}\leqslant 2$, where all $a_i$ are less than 1000, and all $a_i$ have a lowest common multiple greater than 1000.
This is what I have done so far:
If $\frac{1000}{m+1}<a\leqslant{\frac{1000}{m}}$, then there are $m$ multiples of $a$, namely $a,2a,3a,...,ma$ which do not exceed 1000.
We let $k_1$ be the number of $a_i$ in the interval $(\frac{1000}{2},1000]$, $k_2$ the number of $a_i$ in the interval $(\frac{1000}{3},\frac{1000}{2}]$, etc, where $i=1,...,n$.
Then there are $k_1+2k_2+3k_3+...$ integers less than 1000 which are multiples of at least one of the $a_i$.
Why can we say that $k_1+2k_2+3k_3+...<1000$ ?
This is part of the working out of a homework question I have and the fact that I can't explain this step is preventing me from finishing the proof! Thank you so much to whoever helps me out! It is greatly appreciated!
For a positive integer $a<1000$, we define the set $D_a=\{a,2a,\ldots,ma\}$ where $m$ is the biggest integer such that $ma\leq 1000$, that is, $D_a$ is the set of positive multiples of $a$ which do not exceed 1000. Now your hypothesis says that $l.c.m\{a_i,a_j\}>1000$ for $i\neq j$, and this is equivalent to $D_{a_i}\cap D_{a_j}=\emptyset$ for $i\neq j$. Then $D_{a_1}\cup \ldots\cup D_{a_n}$ is a disjoint union which is contained in $\{1,\ldots,1000\}$ (the inclusion is proper because $1$ is not in the union unless $n=1$ and $a_1=1$).
I claim that $k_1+2k_2+\ldots$ is the cardinal of $ D_{a_1}\cup \ldots\cup D_{a_n}$ which was shown to be less than 1000. First, as you noted, if $a\in (\frac{1000}{m+1},\frac{1000}{m}]$ then $|D_a|=m$. Then $$|D_{a_1}\cup \ldots\cup D_{a_n}|=\sum_{i=1}^n|D_{a_i}|=\sum_{m}\sum_{a_i\in (\frac{1000}{m+1},\frac{1000}{m}]}|D_{a_i}|=\sum_mk_mm.$$