無窮級數
請教本題無窮級數的解法，感恩\( \displaystyle \frac{1}{1}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\ldots= \)？
回復 1# rotch 的帖子
\(\begin{align}& 1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\cdots \cdots +\frac{1}{3n-2}+\frac{1}{3n-1}-\frac{2}{3n} \\
& =\left( 1+\frac{1}{2}+\frac{1}{3}-1 \right)+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{2} \right)+\left( \frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{3} \right)+\cdots \cdots +\left( \frac{1}{3n-2}+\frac{1}{3n-1}+\frac{1}{3n}-\frac{1}{n} \right) \\
& =\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{3n} \right)-\left( 1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{n} \right) \\
& =\frac{1}{n+1}+\frac{1}{n+2}+\cdots \cdots +\frac{1}{3n} \\
& \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n+1}+\frac{1}{n+2}+\cdots \cdots +\frac{1}{3n} \\
& =\int_{0}^{2}{\frac{1}{1+x}dx} \\
& =\ln 3 \\
\end{align}\)
回復 2# thepiano 的帖子
感恩您的說明回復 2# thepiano 的帖子
請問您這作答為何我有一台電腦看不到前面,然後另一台電腦又看不到後面 轉成圖檔回復 5# thepiano 的帖子
謝謝您,這樣就很清楚.頁:
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