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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(((x+1)/(x−1)))=ln(((1+(1/x))/(1−((1 )/x))))  ln(1+t)=Σ_(n≥1) (−1)^(n−1) .(t^n /n)  if ∣t∣<1  −ln(1−t)=Σ_(n≥1) (t^n /n)  ⇒ln(((1+(1/x))/(1−(1/x))))=ln(1+(1/x))−ln(1−(1/x))=Σ_(n≥1) (−1)^(n+1) (1/(nx^n ))+Σ_(n≥1) (1/(nx^n ))  =Σ_(n≥1) (1+(−1)^(n+1) ).(1/(nx^n ))=2Σ_(k=1) ^(+∞) (1/((2k−1).x^(2k−1) ))

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

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