纯数学教程 Page 325 例LXVIII (15) 调和级数发散

证明
\begin{equation}
\label{eq:6.31}
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots
\end{equation}发散.

 

假设该级数收敛，则
\begin{equation}
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots=(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots)+\frac{1}{2}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots)
\end{equation}
则
\begin{equation}
\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots=\frac{1}{2}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots)
\end{equation}
且
\begin{equation}
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots=\frac{1}{2}(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots)
\end{equation}
即
\begin{equation}
\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\cdots=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots
\end{equation}
然而$\frac{1}{1}\geq
\frac{1}{2}$,$\frac{1}{3}>\frac{1}{4}$,$\frac{1}{5}>\frac{1}{6}$,$\cdots$因此两者不可能相等.因此假设错误，即
\begin{equation}
\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots
\end{equation}发散.

转载于:https://www.cnblogs.com/yeluqing/archive/2012/11/08/3827762.html