From the power series, $\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - ... + (-1)^{n-1} \frac{x^n}{n} + ... = \ln (1 + x),$ obtain the power series whose sum is $\displaystyle \ln (\frac{1+x}{1-x}).$
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Originally Posted by cgiulz From the power series, $\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - ... + (-1)^{n-1} \frac{x^n}{n} + ... = \ln (1 + x),$ obtain the power series whose sum is $\displaystyle \ln (\frac{1+x}{1-x}).$ (1+x)/(1-x) = 1 + 2x/(1-x) ==> simmilarly as the above one we get: ln(1 - 2x/(1-x)) = 2x/(1-x) - [2x/(1-x)]^2/2 + [2x/(1-x)]^3/3 -.... Tonio
$\displaystyle \ln (1+x)-\ln (1-x)=\ln (1+x)-\ln \big(1+(-x)\big).$
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