How to show that the harmonic function $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions? How can I prove that $H(n) = 1 + \frac{1}{2} + · · · + \frac{1}{n} = O(\log n)$ using simple inequalities on fractions?

For any $n\geq 2$ we have that
$$ \frac{1}{n}<\frac{1}{n}+\frac{1}{3n^3}+\frac{1}{5n^5}+\ldots=\text{arctanh}\frac{1}{n} = \frac{1}{2}\log\frac{n+1}{n-1}\tag{1}$$
hence we may exploit a telescopic sum:
$$\begin{eqnarray*} H_n = 1+\sum_{k=2}^{n}\frac{1}{k} &<& 1+\frac{1}{2}\sum_{k=2}^{n}\left(\log(k+1)-\log(k-1)\right)\\&=&1+\frac{1}{2}\left(\log(n+1)+\log(n)-\log(2)\right)\\&<&\left(1-\frac{\log 2}{2}\right)+\log(n+1).\tag{2} \end{eqnarray*}$$
The inequality
$$ \frac{1}{n}<\log\left(n+\frac{1}{2}\right)-\log\left(n-\frac{1}{2}\right)\tag{3}$$
leads to the sharper (and simpler) bound
$$ \boxed{H_n < \color{red}{ \log(2n+1)}} \tag{4} $$
through the same technique.
