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I've got three inequalities: $\forall n\in\mathbb N:$

$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{1}{2}$$

$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge\frac{7}{12}$$

$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\ge\frac{2}{3}$$

From what I know the LHS converges to something about $0.69$ and each one of them requires the same method, but I can't come up with a proper way to solve it.

Can someone give me a hint?

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    $\begingroup$ You know that you only need to prove one inequality here, and not "three"? $\endgroup$ – Dietrich Burde Oct 7 '17 at 11:31
  • $\begingroup$ As in - if $\frac{1}{n}$+$\frac{1}{n+1}$+...$\frac{1}{2n}$ $\geq$$\frac{2}{3}$, then it will also be $\geq \frac{7}{12}$ and $\geq \frac{1}{2}$ as $\frac{2}{3}>\frac{7}{12}>\frac{1}{2}$ $\endgroup$ – user472341 Oct 7 '17 at 11:34
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    $\begingroup$ Hello! I just wanted to inform you that I’ve made slight edits to the MathJax in your post. Nothing about it was mathematically wrong, per se, so if you feel that the math looked prettier beforehand, feel free to roll it back—and keep contributing to this awesome site! $\ddot\smile$ $\endgroup$ – gen-z ready to perish Oct 7 '17 at 12:14
  • $\begingroup$ In fact it suffice to prove the last inequality using classical inequalities see below $\endgroup$ – Guy Fsone Oct 7 '17 at 17:43
  • $\begingroup$ It's easy to see that $$s_n={1\over n}+{1\over n+1}+\cdots+{1\over2n}\gt{1\over2n}+{1\over2n}+\cdots+{1\over2n}={n+1\over2n}\gt{1\over2}$$ It's also easy to see that $$s_n-s_{n+1}={1\over n}-{1\over2n+1}-{1\over2n+2}=\left({1\over2n}-{1\over2n+1}\right)+\left({1\over2n}-{1\over2n+2}\right)\gt0$$ so that $s_n$ is a decreasing sequence, and thus $$s_n\gt s_{2n}={1\over2n}+\cdots+{1\over3n}+{1\over3n+1}+\cdots+{1\over4n}\gt{n+1\over3n}+{n\over4n}\gt{1\over3}+{1\over4}={7\over12}$$ However, I don't see any similarly easy way to get $s_n\gt{2\over3}$. $\endgroup$ – Barry Cipra Oct 10 '17 at 13:56
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Hint: $$\frac{1}{n}\ge\ln(n+1)-\ln(n)$$

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  • $\begingroup$ Actually, equality seems impossible. $\endgroup$ – Dietrich Burde Oct 7 '17 at 11:34
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    $\begingroup$ @Burde Yes you're right,but it is enough to solve this problem $\endgroup$ – Y.Guo Oct 7 '17 at 11:39
  • $\begingroup$ telescope effect , and you will end up with $\ln(2n)-\ln(n) = \ln(2) \approx 0.693 > \frac{2}{3}$ $\endgroup$ – Ahmad Oct 7 '17 at 12:04
  • $\begingroup$ I got exactly to this point but i wonder if there's any way to estimate the value of $\ln (2)$ other than knowing it $\endgroup$ – Igor Oct 7 '17 at 12:09
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    $\begingroup$ If you know that $e = 2.718\ldots$, then we have $e < 2\sqrt2 = 2^{3/2}$. Therefore $e^{2/3} < 2$, so $\frac23 < \ln2$. $\endgroup$ – Théophile Oct 7 '17 at 12:21
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We know that $$\color{red}{ \frac{x}{x+1}\le\ln(x+1)\le x~~~\forall ~x>0\tag{1}\label{eq}}$$

taking $x=\frac{1}{n+k}~~0\le k\le n$ this lead to $$\ln\left(\frac{1}{n+k}+1\right)\le\frac{1}{n+k}~~~\forall ~~0\le k\le n $$ that is \begin{split} \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} &=&\sum_{k=0}^{n} \frac{1}{n+k} \ge \sum_{k=0}^{n} \ln\left(\frac{1}{n+k}+1\right)\\&=&\sum_{k=0}^{n} \ln\left(n+k+1\right)-\ln\left(n+k\right)\\ &=&\ln\left(2n+1\right)-\ln\left(n\right) \\&= &\ln\left( \frac{2n+1}{n}\right) = \ln\left( 1 +\frac{n+1}{n}\right) \end{split}

Then, Using the left side of $\eqref{eq}$

$$\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n} \ge \ln\left( 1 +\frac{n+1}{n}\right)\ge\frac{n+1}{n+2} \ge \frac{2}{3} $$ Since for all $n\ge 1$ $$\frac{n+1}{n+2} \ge \frac{1+1}{1+2} = \frac{2}{3}$$

Because $x\mapsto \frac{x+1}{x+2}$ is an incresing function with derivative $\frac{1}{(x+2)^2}>0. $

Finally observes that $$\frac{2}{3}>\frac{7}{12}>\frac{1}{2}$$

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