Inegalitate logaritmica

[mihai++]

Demonstrati ca \( 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}<1+[\log_2n],\forall n\in\mathbb{N^*} \).

[Marius Mainea]

Fie \( k=[\log_2n] \)

Atunci \( 2^k\le n<2^{k+1} \) si

\( 1+\frac{1}{2}+...+\frac{1}{n}\le 1+(\frac{1}{2}+\frac{1}{3})+...+(\frac{1}{2^{k-1}}+...\frac{1}{2^k-1})+(\frac{1}{2^k}+...+\frac{1}{n}+...+\frac{1}{2^{k+1}-1})<{1+1+1+....+1}=1+k \)