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Let $n$ be a positive integer and $j=1,2,\ldots, n;$ then I found in an article saying that by using approximations by integrals it is easy to show that $$ \frac{j}{n+1}<1-e^{-(\frac{1}{n}+\cdots + \frac{1}{n-j+1})}<\frac{j}{n+\frac{1}{2}} $$

But I do not see that this is easy and don't know how to use the integral approximations. Any help would be great to proceed further.

Thanks!

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  • $\begingroup$ What article is this from? $\endgroup$ Dec 2, 2016 at 5:24

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Doing elementary manipulations the inequalities can be written as $$ \log\frac{n+1}{n-j+1}<\frac{1}{n-j+1}+\cdots+\frac{1}{n}<\log\frac{n+1/2}{n-j+1/2}. $$ Considering that $\log\frac{a}{b}=\log a-\log b=\int_a^b\frac{dx}{x}$ it seems like a good idea to express this as $$ \int_{n-j+1}^{n+1}\frac{dx}{x}<\frac{1}{n-j+1}+\cdots+\frac{1}{n}<\int_{n-j+1/2}^{n+1/2}\frac{dx}{x}. $$ We have only reworded the problem, but now it becomes obvious if we think in terms of Riemann sums;

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  • $\begingroup$ @SimpleArt I think you will find this pleasingly familiar ;) $\endgroup$
    – user378947
    Dec 2, 2016 at 6:09
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    $\begingroup$ Wow! Thanks, but it is not easy as the authors claimed. $\endgroup$
    – Matt
    Dec 2, 2016 at 14:56

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