A floor function sum involving $\lfloor n/d\rfloor$ and related to sums of divisors

Consider the well known identity from elementary number theory:

$$\sum_{k=1}^n \tau(k)=\sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor,$$ where of course the asymptotic expression for both sides is $n \log n + (2\gamma-1) n + O(\sqrt{n}).$

Some experimentation with Maple seems to show that if $v:=\frac{n}{\log n},$ then $$F(v):=\lim_{n\rightarrow \infty} \frac{\sum_{d=1}^{\lfloor v\rfloor} \left\lfloor \frac{n}{d} \right\rfloor }{\sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor },\quad (1)$$ is increasing towards $1$ from below.

Is it possible to prove what the limit in (1) is, if $v=c\frac{n}{\log n},$ for some $c \in (0,1)$?

migrated from mathoverflow.netJan 25 '17 at 16:17

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• The difference between the top and the bottom should not be far from n(log((log n)/c)-1). Gerhard "Is This Estimate Close Enough?" Paseman, 2017.01.24. – Gerhard Paseman Jan 25 '17 at 6:36
• As @GerhardPaseman says, this seems straightforward: $1-F(\nu)$ is the ratio of the sum from $\nu$ to $n$ divided by the sum from $1$ to $n$. You already have an estimate for the sum from $1$ to $n$, so you just need to prove the sum from $\nu$ to $n$ is of smaller order (which you can do by estimating the terms in the sum). – anthonyquas Jan 25 '17 at 6:53
• Thanks, I will have a go at it and try to formulate an answer. – kodlu Jan 25 '17 at 7:11

The limit can be estimated via (recalling $v=\frac{c n}{\log n}$): $$1-F(v)=\frac{\sum_{d=\lfloor v \rfloor +1}^n \lfloor \frac{n}{d} \rfloor} {\sum_{d=1}^n \lfloor \frac{n}{d} \rfloor} \geq \frac{(n-\frac{cn}{\log n})\log n}{n \log n + (2\gamma-1)n+O(\sqrt{n})} \sim 1-c(\log n)^{-1}$$ by upper bounding the top of the numerator by number of terms times the largest term of the sum.