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For any $n \in \mathbb{N}$, show that: $$\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{2n} < \frac{5}{6}$$

I wrote the sum as $H_{2n} - H_{n}$, where $H_{k} = \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{k}$ (the kth harmonic number). After that, I was searching for inequalities with harmonic numbers, but I didn't find anything useful.

Can you, please, give me a hint? I don't want the full proof. Thank you!

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    $\begingroup$ Hint (I will get you an upper bound, but not the one you are looking for): $$\dfrac{1}{n+1}+\cdots \dfrac{1}{2n} < \underbrace{\dfrac{1}{n+1}+\cdots +\dfrac{1}{n+1}}_{\text{n times}} = \dfrac{n}{n+1}<1$$ The key is to find a comparison with something that is easy to add up. $\endgroup$ – InterstellarProbe Jun 25 '18 at 15:39
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Hint: denoting by $s_n$ the l.h.s., it holds $$ s_n < \int_{n}^{2n} \frac{1}{x} \, dx = \log(2) < \frac{5}{6}\,. $$

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We need to prove that $$\frac{1}{n+1}-\frac{1}{n}+\frac{1}{n+2}-\frac{1}{n}+...+\frac{1}{n+n}-\frac{1}{n}<\frac{5}{6}-1$$ or $$\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{n}{n+n}>\frac{n}{6}.$$ Now, by C-S $$\frac{1}{n+1}+\frac{2}{n+2}+...+\frac{n}{n+n}=$$ $$=\frac{1^2}{1(n+1)}+\frac{2^2}{2(n+2)}+...+\frac{n^2}{n(n+n)}\geq$$ $$\geq\frac{(1+2+...+n)^2}{1(n+1)+2(n+2)+...+n(n+n)}=$$ $$=\frac{\frac{n^2(n+1)^2}{4}}{n(1+2+...+n)+1^2+2^2+...+n^2}=$$ $$=\frac{\frac{n^2(n+1)^2}{4}}{\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}}=\frac{3n(n+1)}{2(5n+1)}>\frac{n}{6}.$$

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