$\sum_{m=1}^{N}\lfloor{\frac{N}{m}\rfloor} \sim N \log N$

Question

Let $d(n)$, the number of divisors of $n$. I am trying to show that $\sum_{n \leq N}d(n) = \#\{(m,k) \in \mathbb{N} \times \mathbb{N}: mk \leq N\} = \sum_{m=1}^{N}\lfloor{\frac{N}{m}\rfloor} \sim N \log N$

I am stuck at showing the asymptotic and how to deal with the floor function. We say $f(n) \sim g(n)$ if $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 1$.

Attempt:

$\lfloor{\frac{N}{m}\rfloor} = \frac{N}{m} + O(1)$. Hence we have $\frac{\frac{N}{m} + O(1)}{N \log N}$.

I am not sure if I can ignore the "Big oh" but think I can as its just a constant. I then tried applying L'hopital by differentiating w.r.t. $N$.

$\frac{1}{m}(N \log N)^{-1} + \frac{N}{m}\Big( \frac{-N \log N + 1}{N^{2}\log^{2}N}\Big)$. But I don't then seem to be able to get this to evaluate out to $1$,I keep ending up with a log factor.

Answer 1

I don't see how you come up with the term $\frac{\frac{N}{m} + O(1)}{N \log N}$. If $\lfloor{\frac{N}{m}\rfloor} = \frac{N}{m} + O(1)$ then $\sum_{m=1}^{N}\lfloor{\frac{N}{m}\rfloor} = N \sum_{m=1}^{N} \frac 1m + O(N) = N \log N + O(N)$.

More precisely: You can estimate $f(N) = \sum_{m=1}^{N}\left\lfloor \frac{N}{m}\right\rfloor$ from above with $$ f(N) \le \sum_{m=1}^{N} \frac Nm = N \sum_{m=1}^{N} \frac 1m \le N( 1 + \log N) $$ and from below with $$ f(N) \ge \sum_{m=1}^{N} \left(\frac Nm -1 \right) = N \sum_{m=1}^{N} \frac 1m - N \ge N \log N - N \, , $$ using $\log N<\sum_{n=1}^{N}\frac{1}{n}<1+\log N.$

It follows that $$ 1 - \frac{1}{\log N} \le \frac{f(N)}{N \log N} \le 1 + \frac{1}{\log N} $$ and therefore $$ \lim_{N \to \infty } \frac{f(N)}{N \log N} = 1 \, . $$

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As @Conrad said, one can use the symmetry of the set $\{(m,k) \in \mathbb{N} \times \mathbb{N}: mk \leq N\}$ to obtain a better asymptotic expression: $$ \sum_{n \leq N}d(n) = 2 \sum_{m=1}^{\lfloor \sqrt N \rfloor } \left\lfloor \frac{N}{m}\right\rfloor - (\lfloor \sqrt N \rfloor)^2 = 2 N \sum_{m=1}^{\lfloor \sqrt N \rfloor } \frac 1m - N + O(\sqrt N) \, . $$ Using the asymptotic expansion $H_N = \log N + \gamma + O(1/N)$ for the harmonic numbers this gives $$ \sum_{n \leq N}d(n) = N \log N + (2\gamma -1) N + O(\sqrt N) \, . $$