What is taylor ln(1+X) ?

The answer to taylor ln(1+X) is sum from n=1 to infinity}(-1)^{n+1 of (x^n)/n

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taylor ln(1+X)

taylor

ln (1 + X)

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

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d X} ( ln ( 1 + X ) ) ( 0 )}{1 !} X + \frac{\frac{d ^{2}}{d X ^{2}} ( ln ( 1 + X ) ) ( 0 )}{2 !} X^{2} + \frac{\frac{d ^{3}}{d X ^{3}} ( ln ( 1 + X ) ) ( 0 )}{3 !} X^{3} + \dots

Evaluate Derivatives

= 0 + \frac{1}{1 !} X + \frac{- 1}{2 !} X^{2} + \frac{2}{3 !} X^{3} + \frac{- 6}{4 !} X^{4} + \frac{24}{5 !} X^{5} + \dots

Refine

= X - \frac{1}{2} X^{2} + \frac{1}{3} X^{3} - \frac{1}{4} X^{4} + \frac{1}{5} X^{5} + \dots

The series sum representation

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

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What is taylor ln(1+X) ?
The answer to taylor ln(1+X) is sum from n=1 to infinity}(-1)^{n+1 of (x^n)/n