$$H(n)=\frac 11+\frac 12+\frac 13+...+\frac 1n$$ I know some approximation to $H(n)$ like below $$\ln(n)=\int_{1}^n \frac{1}{x}dx < H_n< 1+\int_{1}^n \frac{1}{x}dx=1+\ln(n).$$ or $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ but I don't know-how is the proof.
or is there a better approximation for $H(n)$ .
I mean better as 'simple to describe'.
Thanks in advance for any proof,reference,Idea or link...
You already proved that $H_n=\ln n+\mathcal{O}(1)$. Moreover $$ \forall n\in\mathbb{N}^*,\,\frac{1}{n+1}\leqslant\int_n^{n+1}\frac{dx}{x}\leqslant\frac{1}{n} $$ Thus, if $u_n=\frac{1}{n}-\int_n^{n+1}\frac{dx}{x}$ we have $$ 0\leqslant u_n\leqslant \frac{1}{n}-\frac{1}{n+1} $$ and the series $\sum u_n$ converges, we define $\gamma:=\sum_{n=1}^{+\infty}u_n$. However $$ \sum_{k=1}^{n}u_k=H_n-\ln(n+1) $$ so that $H_n=\ln n+\gamma+o(1)$. Moreover, if $v_n=u_n-\ln n$ we have $v_{n+1}-v_n\underset{n\rightarrow +\infty}{\sim}\frac{1}{2n^2}$ so that $$ \sum_{k=n}^{+\infty}{(v_{k+1}-v_k)}\underset{n\rightarrow +\infty}{\sim}\sum_{k=n}^{+\infty}\frac{1}{2k^2}\underset{n\rightarrow +\infty}{\sim}\frac{1}{2n} $$ We finally have $H_n=\ln n+\gamma+\frac{1}{2n}+o\left(\frac{1}{n}\right)$ One can show that $$ H_n \underset{n\rightarrow +\infty}{\sim} \ln n+\gamma+\frac{1}{2n}-\sum_{k=1}^{+\infty}\frac{B_{2k}}{2kn^{2k}} $$ using Euler-Maclaurin formula :
https://en.wikipedia.org/wiki/Harmonic_number https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula