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Thread: Logarithm property

  1. #1
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    Logarithm property

    Can someone help me proof that for any $\displaystyle x \in (-1,1)$ we have $\displaystyle ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}$
    thanks
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    Quote Originally Posted by bram kierkels View Post
    Can someone help me proof that for any $\displaystyle x \in (-1,1)$ we have $\displaystyle ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}$
    thanks
    Hints:
    $\displaystyle \sum\limits_{n = 0}^\infty {x^n } = \frac{1}
    {{1 - x}}\;\& \,\frac{d}
    {{dx}}\ln (1 - x) = ?$
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  3. #3
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    or $\displaystyle -\ln (1-x)=\int_{0}^{x}{\frac{dt}{1-t}},$ and you can use geometric series there to change the sum by the integral.

    note that it's actually for $\displaystyle x\in[-1,1).$
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    Quote Originally Posted by bram kierkels View Post
    Can someone help me proof that for any $\displaystyle x \in (-1,1)$ we have $\displaystyle ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}$
    thanks

    From power series we know that $\displaystyle \frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n\,,\,\,\forall\,x\in\,(-1,1)\,\Longrightarrow -\ln(1-x)=\int\frac{1}{1-x}\,dx=\sum\limits_{n=0}^\infty\int x^n dx$ $\displaystyle =\sum\limits_{n=0}^\infty\frac{x^{n+1}}{n+1}\,,\,\ ,\forall\,x\in\,(-1,1)$
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