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本帖最后由 王守恩 于 2022-5-15 11:00 编辑
\(\displaystyle\sum_{n=1}^{\infty}\frac{(x/a)^{n-1}}{n*2^n}=\frac{a*\log(2a/(2a-x))}{x}=\frac{\log(2/(2-x/a))}{x/a}=\frac{\log(2)-\log(2-x/a)}{x/a}\)
\(-2a≤x≤2a-1\ \ \ \ x≠0\ \ \ \ a=1,2,3,4,5,6,7,....\)
\(当x=-2a时,\)
\(\displaystyle\sum_{n=1}^{\infty}\frac{(x/a)^{n-1}}{n*2^n}=\frac{\log(2)-\log(2+2)}{2}=\frac{\log(2)}{2}=0.34657359027997265471\)
\(中间在慢慢长大\)
\(最后,当x=2a-1时,\)
\(\displaystyle\sum_{n=1}^{\infty}\frac{(x/a)^{n-1}}{n*2^n}=\frac{\log(2)+\log(a)}{2}=\infty\) |
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