I am studying power series now and I'm confused about those $\ln$ sums:
$$\ln(1+x) = \sum_{n=1}^{\infty} {{(-1)^{n+1} x^{n} \over n}} \text{ and } \ln(1-x) = \sum_{n=1}^{\infty} {x^n \over n}$$
since $|x| < 1$;
Say $x = 0.9$, then $\ln(1-0.9) < 0$. In this case, how we're going to get a negative number in the power series since the expansion of $\ln(1-x)$ has only positive numbers.
See this:
Prove that $$\ln\sqrt{1+x\over1-x}=\sum_{n=0}^{\infty}{x^{2n+1}\over 2n+1}$$
$\ln\sqrt{1+x\over1-x} = {1 \over 2}[ \ln(1+x) - \ln(1-x) ] = {1 \over 2} [\sum_{n=1}^{\infty} {{(-1)^{n+1} x^{n} \over n}} - \sum_{n=1}^{\infty} {x^n \over n}]=$
$= {1 \over 2} \{[x - {x²\over2}+{x³ \over3}-\ldots] - [x+{x²\over2} + {x³\over3} \ldots] \}$ This make me get rid of all the odd terms of the equation, and it does not make me get the equality I'm looking for. If only I could have all the terms in the second term to be negative then all the even terms would vanish and I would get the result I'm interested.