What is the integer part of the sum of the reciprocals of the first $2017$ positive integers?

Question

Finding value of $\displaystyle \bigg\lfloor 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots \cdots +\frac{1}{2017}\bigg\rfloor $, where $\lfloor x \rfloor = x - \{x\}$.

Attempt: taking $\displaystyle f(x) = \frac{1}{x}$, then $\displaystyle \int^{2018}_{1}\frac{1}{x}dx=\ln(2018) <\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots \cdots +\frac{1}{2017}$

could some help how i find upper bound and integer part of $\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots \cdots +\frac{1}{2017}$

thanks in advanced

Answer 1

A very good approximation to $$H_n=\sum_{j=1}^n \frac{1}{j}$$ is $$\ln(n)+\gamma+\frac{1}{2n}$$ This gives about $8.19$, so the integer part is $8$

Answer 2

In this case, instead of an approximation I would rather use an actual bound to be sure that we don't have any residual error that could cause the floor function to step up or down.

This tells us that : $$ 8.18682 < \log(2017) + \gamma + \frac{1}{2\cdot (2017+1)} < \sum_{i=1}^{2017} \frac{1}{i} < \log(2017) + \gamma + \frac{1}{2\cdot 2017} < 8.18684 $$