<s>[INST] 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)?
 [/INST] 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.
$$\ln(x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n} \implies \ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x)^n}{n}$$
$$\frac{d}{dx}\ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{n(x)^{n-1}}{n} \implies \frac{1}{1+x} = \sum_{n=1}^{\infty} (-1)^{n+1}x^{n-1}$$
$$\frac{1}{1+(-x)} = \sum_{n=1}^{\infty} (-1)^{n+1}(-x)^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} (-1)^{n+1}(-1)^{n-1}x^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$
$$ \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2(n-1)} \implies \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$
$$\frac{x}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-1}$$
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