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Find $c > 0$ such that for any positive integer $n$, $$\sum_{k = 1}^{n}\frac{1}{k} \geq c \log n.$$ As I am now confused with the question written here Showing that $\sum_{i=1}^n \frac{1}{i} \geq \log{n}$ , could anyone give me a hint about the proof?

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closed as unclear what you're asking by Simply Beautiful Art, Arnaldo, Daniel W. Farlow, kingW3, Jack D'Aurizio Jun 6 '17 at 14:04

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  • $\begingroup$ what $c$ is that? $\endgroup$ – RGS Jun 6 '17 at 11:15
  • $\begingroup$ @RSerrao a constant. $\endgroup$ – Intuition Jun 6 '17 at 11:17
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    $\begingroup$ You're asking why you're asking this question, and not another one? Seriously? $\endgroup$ – Professor Vector Jun 6 '17 at 11:23
  • $\begingroup$ So Sorry I got it @ProfessorVector. but because the constant in my problem the condition on it is $c > 0$ only so that the other inequality may not be satisfied. $\endgroup$ – Intuition Jun 6 '17 at 11:24
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Hint. Note that for any positive integer $k$, $$\ln(k+1)-\ln(k)=\int_{k}^{k+1}\frac{1}{x}\, dx\leq \frac{1}{k}.$$ Then $$\ln(n+1)=\sum_{k=1}^n(\ln(k+1)-\ln(k))\leq \sum_{k=1}^n\frac{1}{k}.$$

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  • $\begingroup$ I do not know how to use your hint. $\endgroup$ – Intuition Jun 6 '17 at 12:52
  • $\begingroup$ See my edited answer. Which constant $c$ are you going to choose? $\endgroup$ – Robert Z Jun 6 '17 at 12:59
  • $\begingroup$ the constant c = 1 $\endgroup$ – Intuition Jun 6 '17 at 15:06
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    $\begingroup$ @weakmathematician Yes, actually any $c\in (0,1]$ is fine. $\endgroup$ – Robert Z Jun 6 '17 at 15:10
  • $\begingroup$ @weakmathematician Please edit your question and improve it. Say something like "Find $c>0$ such that for any positive integer $n$..." $\endgroup$ – Robert Z Jun 6 '17 at 15:19
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For any $x\in(0,1)$ we have $\log(1+x)\leq x$, hence $$ H_n=\sum_{k=1}^{n}\frac{1}{k}\color{red}{\geq} \sum_{k=1}^{n}\log\left(1+\frac{1}{k}\right) = \sum_{k=1}^{n}\left[\log(k+1)-\log k\right] = \log(n+1).$$ We also have $x\leq \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)$, from which: $$ H_n \leq 1+\frac{1}{2}\sum_{k=2}^{n}\left[\log\left(1+\frac{1}{k}\right)-\log\left(1-\frac{1}{k}\right)\right]\color{red}{\leq }\log(n+1)+\left(1-\frac{\log 2}{2}\right)$$ for any $n\geq 2$.

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