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Let $a\geq 1$. Must there exist some $b>a$ such that \[\sum_{a\leq n\leq b}\frac{1}{n}=\frac{r_1}{s_1}\textrm{ and }\sum_{a\leq n\leq b+1}\frac{1}{n}=\frac{r_2}{s_2},\] with $(r_i,s_i)=1$ and $s_2<s_1$? If so, how does this $b(a)$ grow with $a$?
For example, \[\sum_{3\leq n\leq 5}\frac{1}{n} = \frac{47}{60}\textrm{ and }\sum_{3\leq n\leq 6}\frac{1}{n}=\frac{19}{20}.\]

The smallest $b$ for each $a$ are listed at A375081 at the OEIS.

This was resolved in the affirmative by van Doorn [vD24], who proved $b=b(a)$ always exists, and in fact $b(a) \ll a$. Indeed, if $a\in (3^k,3^{k+1}]$ then one can take $b=2\cdot 3^{k+1}-1$. van Doorn also proves that $b(a)>a+(1/2-o(1))\log a$, and considers various generalisations of the original problem.

It seems likely that $b(a)\leq (1+o(1))a$, and perhaps even $b(a)\leq a+(\log a)^{O(1)}$.

Additional thanks to: Ralf Stephan and Wouter van Doorn