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Math Help - Series

  1. #1
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    Series

    Show by integrating the series for 1/(1 + x), that if |x| < 1, then

    \ln (1 + x) = {\Sigma_{n=1}^{\infty}}\frac {-1^{(n+1)}}{n} x^n
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  2. #2
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    isn't that use Taylor expansion series?
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  3. #3
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    Consider the suggested series and integrate both sides (the RHS term-by-term and let u = 1+x for the LHS):

    \displaystyle \frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^nx^n \Rightarrow \int\frac{1}{1+x}\;{dx} = \int\sum_{n=0}^{\infty}(-1)^nx^n \;{dx} \Rightarrow \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}.
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