How do you find the power series of ln(1+x)?

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How do you find the power series of ln(1+x)?

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Answer 1 · 2017-02-06T00:33:23

$ln (1 + x) = \infty \sum n = 0 {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}$ $ln(1+x) = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)$

with radius of convergence $R = 1$ $R=1$

Explanation:

Start from:

$ln (1 + x) = \int_{0}^{x} \frac{d t}{1 + t}$ $ln(1+x) = int_0^x (dt)/(1+t)$

Now the integrand function is the sum of a geometric series of ratio $- t$ $-t$ :

$\frac{1}{1 + t} = \infty \sum n = 0 {(- 1)}^{n} t^{n}$ $1/(1+t) = sum_(n=0)^oo (-1)^nt^n$

so:

$ln (1 + x) = \int_{0}^{x} \infty \sum n = 0 {(- 1)}^{n} t^{n}$ $ln(1+x) = int_0^x sum_(n=0)^oo (-1)^nt^n$

This series has radius of convergence $R = 1$ $R=1$ , so in the interval $x \in (- 1, 1)$ $x in (-1,1)$ we can integrate term by term:

$ln (1 + x) = \infty \sum n = 0 \int_{0}^{x} {(- 1)}^{n} t^{n} d t = \infty \sum n = 0 {(- 1)}^{n} \frac{x^{n + 1}}{n + 1}$ $ln(1+x) = sum_(n=0)^oo int_0^x (-1)^nt^ndt = sum_(n=0)^oo (-1)^n x^(n+1)/(n+1)$

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How do you find the power series of ln(1+x)?

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