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What is the largest possible numerator when put in reduced form over all sums of the form $$\sum_{k=1}^n\frac{c(k)}{k}$$ where $c(k)\in\{-1, 0, 1\}$? An easy bound is to consider what happens when we don't reduce the fraction. Then we have that if $L$ is the $\text{lcm}$ over all $k$ such that $c(k)\neq 0$, then $$\Big|\text{num}\left(\sum_{k=1}^n\frac{c(k)}{k}\right)\Big|\le \Big|L\sum_{k=1}^n\frac{c(k)}{k}\Big|\le L\sum_{k=1}^{n}\frac{|c(k)|}{k}\le H_n\text{lcm}(1,\dots,n)$$ where $H_n$ is the $n$-th harmonic number. For some $n$, this bound is the best achievable, which can be seen when the denominator of $H_n$ truly is $\text{lcm}(1,\dots, n)$. Can any improvements be made or can this be shown to be the best bound for certain subsets of numbers? Another useful result would be the asymptotics of this, even if for only certain subsets.
This problem came from considering this problem involving egyptian fractions.

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