I am given the Maclaurin series for $ \ln(1+x) $, namely $ \ln(1 + x) = \sum\limits_{n=1}^\infty \dfrac{(-1)^{n+1}x^n}{n} $ , $ -1 < x \leq 1 $.
I now need to prove $$ \ln\left(\frac{1}{2-x^2}\right) =\sum_{n=1}^\infty \frac{(-1)^n(1-x^2)^n}{n} \ , \ |x| < \sqrt{2}. $$
My attempt:
I started by rewriting the LHS:
$$ \ln\left(\frac{1}{2-x^2}\right) = - \ln(2-x^2) = -\ln\left(2\left(1 - \frac{x^2}{2}\right)\right) = -\left(\ln2 + \ln\left(1-\frac{x^2}{2}\right)\right) $$
Then I substituted $ u = -\dfrac{x^2}{2} $ to get $ -(\ln 2 + \ln(1 + u)) $.
Next I put in the Maclaurin series for $ \ln 2 = \sum\limits_{n=1}^\infty \dfrac{(-1)^{n+1}}{n} $ and $ \ln(1 + u) $ (given):
$$ -(\ln 2 + \ln(1 + u)) = -\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} + \sum_{n=1}^\infty \frac{(-1)^{n+1}u^n}{n}\right) $$
I added the series together, factored out the $ (-1)^{n+1} $ and converted it to $ (-1)^n $ using the minus sign outside the series:
$$ -(\ln 2 + \ln(1 + u)) = \sum_{n=1}^\infty \frac{(-1)^{n}(1 + u^n)}{n} $$
Finally, I unsubstituted u:
$$ \ln\left(\frac{1}{2-x^2}\right) = \sum\limits_{n=1}^\infty \frac{(-1)^{n}(1 + (-\frac{x^2}{2})^n)}{n} $$
Some graphing shows that this series does indeed converge to the LHS, but I am stuck on how to do the algebra and convert this to the form shown in the RHS. WolframAlpha gives the RHS but does not show its work.
It is simpler than you think:
$$-\ln(2-x^2)=-\ln(1+1-x^2)=\sum_{n=1}^{\infty}\frac{(-1)^{n}(1-x^2)^n}{n}$$ valid whenever $|1-x^2|<1$, i.e. $|x|<\sqrt{2}$.