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Let's define $\sigma(n,k) = \sum_{d|n,d\leq k} d$. Notice $\sigma(n,n)=\sigma(n)$, the standard sum of divisor function.

It's a standard exercise to show that $$ \sigma(n) \leq n (\ln n + 1) $$

for some constant $c$.

Is there any known result on the asymptotic when we want to compute the sum of divisors of $n$ up to $k$? Do we expect

$$ \sigma(n,k) \leq k f(n) $$

where $f(n)$ is some small function depending on $n$? Is there any study of this sum? It be great if I can find references, as I need to extend it for something for abelian groups.

It's not hard to see a few partial results. Say if $n$ is a power of a prime, we can let $f(n)=\log_2 n$. If $1,\ldots,k$ are divisors of $n$, then $\sigma(n,k) = {k+1 \choose 2}$. However this implies $n$ is at least the product of all prime numbers between $1$ and $k$, hence $n\sim e^k$. So this seems $f(n) = \ln n$ is sufficient.

So one would conjecture that $\sigma(n,k)\leq k c(\ln n + 1)$ where $c>0$ is some constant.

If one only care about the asymptotic, then $f(n)=o(n^{\epsilon})$ for any $\epsilon>0$. This comes from the number of divisors are at most $n^\epsilon$, and each divisor in the sum can contribute at most $k$.

Chao Xu
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    Good question!—I can't immediately prove (nor disprove) your cojecture. Remark 1: it suffices to consider the case where $n$ divides $\mathop{\rm lcm}[1,2,\dots,k]$, since replacing $n$ by $\gcd(n,\mathop{\rm lcm}[1,2,\dots,k])$ doesn't change $\sigma(n,k)$. Remark 2: the conjecture can be restated as follows. Let $S$ be a set of positive integers closed under taking divisors; then the conjecture is that $\sum_{d\in S} d \le \max S \cdot c(\ln\mathop{\rm lcm}S + 1)$. – Greg Martin May 21 '15 at 06:30

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