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Question Number 20561 by mondodotto@gmail.com last updated on 28/Aug/17

Answered by mind is power last updated on 07/Nov/19

$$\ln \left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)=\ln \left(\frac{1+\frac{1}{\mathrm{x}}}{1-\frac{1}{\mathrm{x}}}\right)$$$$\ln (1+\mathrm{t})=\sum _{\mathrm{n}⩾1}{(-1)}^{\mathrm{n}-1}.\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}}\mathrm{if}\mid \mathrm{t}\mid <1$$$$-\ln (1-\mathrm{t})=\sum _{\mathrm{n}⩾1}\frac{{\mathrm{t}}^{\mathrm{n}}}{\mathrm{n}}$$$$\Rightarrow \ln \left(\frac{1+\frac{1}{\mathrm{x}}}{1-\frac{1}{\mathrm{x}}}\right)=\ln (1+\frac{1}{\mathrm{x}})-\ln (1-\frac{1}{\mathrm{x}})=\sum _{\mathrm{n}⩾1}{(-1)}^{\mathrm{n}+1}\frac{1}{{\mathrm{nx}}^{\mathrm{n}}}+\sum _{\mathrm{n}⩾1}\frac{1}{{\mathrm{nx}}^{\mathrm{n}}}$$$$=\sum _{\mathrm{n}⩾1}(1+{(-1)}^{\mathrm{n}+1}).\frac{1}{{\mathrm{nx}}^{\mathrm{n}}}=2\underset{\mathrm{k}=1}{\sum ^{+\mathrm{\infty }}}\frac{1}{(2\mathrm{k}-1).{\mathrm{x}}^{2\mathrm{k}-1}}$$$$$$

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