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Question Number 172545 by SANOGO last updated on 28/Jun/22

calcul  Σ_(n=1) ^(+oo) (((−1)^n )/n)x^n

$${calcul} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+{oo}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{x}^{{n}} \\ $$

Commented by mr W last updated on 28/Jun/22

Σ_(n=1) ^∞ (((−1)^(n+1) x^n )/n)=ln (1+x)  −Σ_(n=1) ^∞ (((−1)^n x^n )/n)=ln (1+x)  ⇒Σ_(n=1) ^∞ (((−1)^n x^n )/n)=ln (1/(1+x))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{{n}} }{{n}}=\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$$$−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} }{{n}}=\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$$$\Rightarrow\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} }{{n}}=\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{1}+{x}} \\ $$

Answered by Jamshidbek last updated on 28/Jun/22

(1/(1+x))=1−x+x^2 −x^3 +...=Σ_(n=0) ^∞ (−1)^n ∙x^n   (1/(1+x))=Σ_(n=0) ^∞ (−1)^n ∙x^n      ∣x∣<1  integrate both side  ln(1+x)=Σ_(n=0) ^∞ (−1)^n ∙(x^(n+1) /(n+1))=Σ_(n=1) ^∞ (−1)^(n−1) ∙(x^n /n)  −ln(1+x)=Σ_(n=1) ^∞ (−1)^n ∙(x^n /n)    ∣x∣<1  Telegram:@math_undergraduate

$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}=\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} +...=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \centerdot\mathrm{x}^{\mathrm{n}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \centerdot\mathrm{x}^{\mathrm{n}} \:\:\:\:\:\mid\mathrm{x}\mid<\mathrm{1} \\ $$$$\mathrm{integrate}\:\mathrm{both}\:\mathrm{side} \\ $$$$\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \centerdot\frac{\mathrm{x}^{\mathrm{n}+\mathrm{1}} }{\mathrm{n}+\mathrm{1}}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} \centerdot\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}} \\ $$$$−\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \centerdot\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}}\:\:\:\:\mid\mathrm{x}\mid<\mathrm{1} \\ $$$$\mathrm{Telegram}:@\mathrm{math\_undergraduate} \\ $$$$ \\ $$

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