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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

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    $\begingroup$ From $$\frac{1}{k+1} \le \int_{k}^{k+1} \frac{dt}{t} \le \frac{1}{k}$$ you get $$\sum_{k = 1}^n \frac{1}{k+1} \le \int_{1}^{n+1} \frac{dt}{t} \le \sum_{k = 1}^n \frac{1}{k}$$ So $$ \underbrace{\int_{1}^{n+1} \frac{dt}{t}}_{\ln(n+1)} \le \sum_{k = 1}^n \frac{1}{k} \le 1 + \underbrace{\int_{1}^{n} \frac{dt}{t}}_{\ln(n)} $$ Unfortunately, this approximation isn't sharp enough, because $\ln(2018) < 8 < \ln(2017)+1$. $\endgroup$ – Joel Cohen Mar 27 '17 at 11:12
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    $\begingroup$ But the slightly better approximation $H_n\approx \ln(n)+\gamma$ is sufficient here. $\endgroup$ – Peter Mar 27 '17 at 11:15
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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 $8$

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  • $\begingroup$ If you are extremely unlucky that the approximation is very very near to an integer, the formula might give the wrong integer part for some $n$, but here everything is alright. The error is about $2\cdot 10^{-8}$, which is far smaller than what we would need. $\endgroup$ – Peter Mar 27 '17 at 11:08
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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 : $$ 8.18682 < \log(2017) + \gamma + \frac{1}{2\cdot (2017+1)} < \sum_{i=1}^{2017} \frac{1}{i} < \log(2017) + \gamma + \frac{1}{2\cdot 2017} < 8.18684 $$

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