Find Taylor expansion of $f(x)=\ln{1-x^2\over 1+x^2}$, and then find radius of convergence. Find Taylor expansion of $f(x)=\ln{1-x^2\over 1+x^2}$, and then find radius of convergence. My problem here is the Taylor series. Computing the few first derivative is possible, but I can't seem to have caught any pattern I can formulate elegantly. Maybe I should only make an observation about the derivatives without using a formula? I could really use any help here.
Edit: I arrived at: $$f(x)=-2\sum_{n=0}^{\infty}{x^{2+4n}\over 2n+1}$$. Using the ratio test: $$\lim\limits_{n\to \infty}|{a_{n+1}\over a_n}|=\lim\limits_{n\to \infty}|{{x^2x^{4n}x^4\over 2n+3}\over {x^2x^{4n}\over 2n+1}}|=\lim\limits_{n\to \infty}|{x^4(2n+1)\over 2n+3}|=|x^4|$$. $|x^4|<1\iff |x|<1 \Rightarrow R=1$.

Use the Taylor series for $\ln(1+u)$ and $\ln(1-u)$ and subsitute  $x^2$ to $u$:
\begin{align*}
\ln(1-x^2)&=-x^2-\dfrac{x^4}2-\dfrac{x^6}3-\dotsm - \dfrac{x^{2n}}{n}-\dotsm\\
\ln(1+x^2)&=x^2-\dfrac{x^4}2+\dfrac{x^6}3-\dotsm +(-1)^{n-1} \dfrac{x^{2n}}{n}+\dotsm
\end{align*}
We obtain:
$$\ln\dfrac{1-x^2}{1+x^2}=\ln(1-x^2)-\ln(1+x^2)=-2\Bigl(x^2+\frac{x^6}3+\dots+\dfrac{x^{2(2k+1)}}{2k+1}+\dotsm\Bigr)$$
Radius of convergence: The function is defined on $(-1,1)$, and the radius of convergence of the Taylor series of $\ln(1\pm u)$ is $1$. Hence the radius of convergence for the Taylor series of $f(x)$ is $1$.
