调和级数

$$\sum _{n=1}^{\infty }\frac{1}{n}$$ $$\begin{array}{rl}{S}_{2n}-S_n& =\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}\\ & >n\cdot \frac{1}{2n}\\ & =\frac{1}{2}\end{array}$$

反设收敛

$$⟹\underset{n\to \infty }{\lim }(S_{2n}-S_n)=S-S=0$$

矛盾，因此级数不收敛

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另法：不等式放缩

$$x\ge \ln (1+x)$$ $$\frac{1}{k}\ge \ln \left(\frac{1+k}{k}\right)$$ $$S_n\ge \ln 2+\ln \frac{3}{2}+\ln \frac{4}{3}+\cdots =\ln (n+1)\to +\infty$$

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另法：中值不等式

$$\ln (k+1)-\ln (k)=\frac{1}{\theta }<\frac{1}{k}$$