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Math Help - Logarithm property

  1. #1
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    Logarithm property

    Can someone help me proof that for any x \in (-1,1) we have  ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}
    thanks
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  2. #2
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    Quote Originally Posted by bram kierkels View Post
    Can someone help me proof that for any x \in (-1,1) we have  ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}
    thanks
    Hints:
    \sum\limits_{n = 0}^\infty  {x^n }  = \frac{1}<br />
{{1 - x}}\;\& \,\frac{d}<br />
{{dx}}\ln (1 - x) = ?
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  3. #3
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    or -\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 x\in[-1,1).
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    Quote Originally Posted by bram kierkels View Post
    Can someone help me proof that for any x \in (-1,1) we have  ln(1-x) = -\sum_{n=1}^{\infty}{\frac{x^n}{n}}
    thanks

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