# 無窮級數

## 無窮級數

$$\displaystyle \frac{1}{1}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\ldots=$$？

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## 回復 1# rotch 的帖子

\begin{align} & 1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\cdots \cdots +\frac{1}{3n-2}+\frac{1}{3n-1}-\frac{2}{3n} \\ & =\left( 1+\frac{1}{2}+\frac{1}{3}-1 \right)+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{2} \right)+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{3} \right)+\cdots \cdots +\left( \frac{1}{3n-2}+\frac{1}{3n-1}+\frac{1}{3n}-\frac{1}{n} \right) \\ & =\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{3n} \right)-\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n} \right) \\ & =\frac{1}{n+1}+\frac{1}{n+2}+\cdots \cdots +\frac{1}{3n} \\ & \\ & \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n+1}+\frac{1}{n+2}+\cdots \cdots +\frac{1}{3n} \\ & =\int_{0}^{2}{\frac{1}{1+x}dx} \\ & =\ln 3 \\ \end{align}

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## 回復 2# thepiano 的帖子

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#### 附件

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2017-4-7 18:05

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