Prove that $\sum_{k = 1}^{n}\frac{1}{k} \geq c\space \log n$. [closed]

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?

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

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• what $c$ is that? – RGS Jun 6 '17 at 11:15
• @RSerrao a constant. – Intuition Jun 6 '17 at 11:17
• You're asking why you're asking this question, and not another one? Seriously? – Professor Vector Jun 6 '17 at 11:23
• 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. – Intuition Jun 6 '17 at 11:24

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 $$\ln(n+1)=\sum_{k=1}^n(\ln(k+1)-\ln(k))\leq \sum_{k=1}^n\frac{1}{k}.$$
• See my edited answer. Which constant $c$ are you going to choose? – Robert Z Jun 6 '17 at 12:59
• @weakmathematician Yes, actually any $c\in (0,1]$ is fine. – Robert Z Jun 6 '17 at 15:10
• @weakmathematician Please edit your question and improve it. Say something like "Find $c>0$ such that for any positive integer $n$..." – Robert Z Jun 6 '17 at 15:19
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 $n\geq 2$.