一个数项级数发散的证明

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问题    证明级数$$\sum_{n=1}^{\infty} \frac{\cos( \frac{\pi} { 2} \log n)}{n}$$发散.

证明    取$$m=\left[e^{4N}+1\right],n=\left[e^{4N+1}\right]$$

那么当$x\in[m,n]$时,函数$\frac{\cos\left(\frac{\pi}{2}\log x\right)}{x}$递减,考虑

\begin{align*}\sum_{k=m}^{n-1}\frac{\cos\left(\frac{\pi}{2}\log k\right)}{k}&\geq\sum_{k=m}^{n-1}\int_{k}^{k+1}\frac{\cos\left(\frac{\pi}{2}\log x\right)}{x}{\rm d}x\\&=\int_{m}^{n}\frac{\cos\left(\frac{\pi}{2}\log x\right)}{x}{\rm d}x\\&=\frac{2}{\pi}\left.\sin\left(\frac{\pi}{2}\log x\right)\right|_{m}^{n}\tag{1}\end{align*}

注意到$$\log n>\log\left(e^{4N+1}-1\right),\log m<\log\left(e^{4N}+1\right)$$

带入(1)式得\begin{align*}\sum_{k=m}^{n-1}\frac{\cos\left(\frac{\pi}{2}\log k\right)}{k}&\geq\frac{2}{\pi}\left[\sin\left(\frac{\pi}{2}\log\left(e^{4N+1}-1\right)\right)-\sin\left(\frac{\pi}{2}\log\left(e^{4N}+1\right)\right)\right]\\&=\frac{2}{\pi}\left[\sin\left(\frac{\pi}{2}+\frac{\pi}{2}\log\left(1-\frac{1}{e^{4N+1}}\right)\right)-\sin\left(\frac{\pi}{2}\log\left(1+\frac{1}{e^{4N}}\right)\right)\right]\\&\to\frac{2}{\pi},(N\to\infty)\end{align*}

由Cauchy收敛原理可得原级数发散

一个数项级数发散的证明

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原文地址：http://www.cnblogs.com/xiaoxixi/p/4211254.html