calcul-n-1-oo-1-n-n-x-n-

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

c a l c u l

\underset{n = 1}{\sum^{+ o o}} \frac{{(- 1)}^{n}}{n} x^{n}

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

\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} x^{n}}{n} = \ln (1 + x)

- \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{n}}{n} = \ln (1 + x)

\Rightarrow \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{n}}{n} = \ln \frac{1}{1 + x}

Answered by Jamshidbek last updated on 28/Jun/22

\frac{1}{1 + x} = 1 - x + x^{2} - x^{3} + \dots = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \cdot x^{n}

\frac{1}{1 + x} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \cdot x^{n} ∣ x ∣< 1

integrate both side

\ln (1 + x) = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \cdot \frac{x^{n + 1}}{n + 1} = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n - 1} \cdot \frac{x^{n}}{n}

- \ln (1 + x) = \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \cdot \frac{x^{n}}{n} ∣ x ∣< 1

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