Sum with floor of harmonic series

Question

How can I prove the following? $$\sum_{i=1}^{n} \left\lfloor \frac{n}{i} \right\rfloor \sim n \sum_{i=1}^{n}\frac{1}{i}$$

Answer 1

Denote $S = \sum_{i=1}^{n} \frac{n}{i}$. Use

$$\sum_{i=1}^{n} \frac{n}{i}-1 \le \sum_{i=1}^{n} \left\lfloor \frac{n}{i} \right\rfloor \le \sum_{i=1}^{n} \frac{n}{i} $$

$$\left(\sum_{i=1}^{n} \frac{n}{i}\right)-n \le \sum_{i=1}^{n} \left\lfloor \frac{n}{i} \right\rfloor \le \sum_{i=1}^{n} \frac{n}{i} $$ Note that $$\dfrac{\left(\sum_{i=1}^{n} \frac{n}{i}\right)-n}{\sum_{i=1}^{n} \frac{n}{i}} \rightarrow 1 $$

So you can get required by squeeze theorem.