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Question Number 181570 by SANOGO last updated on 26/Nov/22

prove that:xε]−1,1[  Σ_(n=1) ^(+oo) (x^n /n)=−ln(1−x)

$$\left.{prove}\:{that}:{x}\epsilon\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}\frac{{x}^{{n}} }{{n}}=−{ln}\left(\mathrm{1}−{x}\right) \\ $$

Answered by mr W last updated on 27/Nov/22

for −1<x<1:  1+x+x^2 +...=Σ_(n=1) ^∞ x^(n−1) =(1/(1−x))  Σ_(n=1) ^∞ ∫_0 ^x x^(n−1) dx=∫_0 ^x (1/(1−x))dx  Σ_(n=1) ^∞ (x^n /n)=−ln (1−x) ✓

$${for}\:−\mathrm{1}<{x}<\mathrm{1}: \\ $$$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +...=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{{x}} {x}^{{n}−\mathrm{1}} {dx}=\int_{\mathrm{0}} ^{{x}} \frac{\mathrm{1}}{\mathrm{1}−{x}}{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}}=−\mathrm{ln}\:\left(\mathrm{1}−{x}\right)\:\checkmark \\ $$

Commented by SANOGO last updated on 27/Nov/22

thank you

$${thank}\:{you} \\ $$

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