证明：

$\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dots+\dfrac{1}{2^k-1}$

$<\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{8}+\dots+\dfrac{1}{2^k-1}$

$=1+1+1+\dots+1$

$∴1+\dots+\dfrac{1}{2^k-1}<k=\log2^k$

$∴1+\dots+\dfrac{1}{n}<\log n$