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

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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.

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  • $\begingroup$ I like your answer. I think it is humorous that you chose to "Denote $S = \sum_{i=1}^{n} \frac{n}{i}" and proceeded to not use this anywhere else. $\endgroup$ – abiessu Jan 25 '18 at 3:26

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