What is the interval of convergence of the Taylor series $f(x) = \ln(1-x^2)$ centered at $x = 0$?
For $f(x) = \ln(1-x^2) = \ln(1-x) + \ln(1+x) $ I've found that the nth derivative for $x=0$ is $-(n-1)! + (-1)^{(n+1)}*(n-1)!$
Thus, the Taylor series centered at $x = 0$ is $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}(x-0)^n$ = $\sum_{n=0}^\infty \frac{-(n-1)! + (-1)^{(n+1)}*(n-1)!}{n!}(x)^n$ = $\sum_{n=1}^\infty \frac{-1+ (-1)^{(n+1)}}{n}(x)^n$ = $0-x^2 +0 - \frac{x^4}{4}+\ldots$
Which is equal to $\sum_{n=1}^\infty \frac{-1}{2^{2n-1}}(x)^{2n}$, when $n$ is a even number and $=0$ when $n$ is an odd number.
Because of this, we can't apply the ratio test.
Then I tried to see if the series is absolutely convergent by putting it inside modulo. The resulting series is $\sum_{n=0}^\infty \frac{1}{2^{2n-1}}(x)^{2n}$
I applied the ratio test here:
$\lim_{n\to\infty} \frac{x_n}{x_{n+1}} $ = $\lim_{n\to\infty} \frac{x^{2n}}{2^{2n-1}} \frac{2^{2n-3}}{x^{2n+2}} = \lim_{n\to\infty} \frac{1}{4x^2} <1 $ for any value of $x$. Thus the series is never absolutely convergent.
Because of this, I said that the interval of convergence is $\emptyset$.
Is this alright?
Since$$\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\cdots$$when $|x|<1$ and since $|x|<1\iff|x^2|<1$, you have$$\log(1-x^2)=-x^2-\frac{x^4}2-\frac{x^6}3-\cdots\tag1$$when $|x|<1$. It's easy to see that the series $(1)$ diverges when $|x|>1$ and therefore the radius of convergence is $1$.