Finding Taylor Series from existing Series If the Taylor Series of $\ln(x)$ is known:
$$\ln(x) = (x-1) -\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3-\frac{1}{4}(x-1)^4+\frac{1}{5}(x-1)^5-...$$
Can one find the Taylor series of
$$f(x)= \frac{x}{1-x^2}$$
by manipulating the Taylor series of ln(x)?

This isn't really an answer on its own, but I'll add this here in case you want to see how to get the series in detail.
$$\ln(x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n} \implies \ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x)^n}{n}$$
$$\frac{d}{dx}\ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{n(x)^{n-1}}{n} \implies \frac{1}{1+x} = \sum_{n=1}^{\infty} (-1)^{n+1}x^{n-1}$$
$$\frac{1}{1+(-x)} = \sum_{n=1}^{\infty} (-1)^{n+1}(-x)^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} (-1)^{n+1}(-1)^{n-1}x^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$
$$ \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2(n-1)} \implies \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$
$$\frac{x}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-1}$$
