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

Question

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?

Answer 1

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 $$\ln(n+1)=\sum_{k=1}^n(\ln(k+1)-\ln(k))\leq \sum_{k=1}^n\frac{1}{k}.$$

Answer 2

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 $n\geq 2$.