<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] Make the substitution $x = 1 - u^{2}$ in order to obtain:
\begin{align*}
\ln(1 - u^{2}) = -u^{2} - \frac{u^{4}}{2} - \frac{u^{6}}{3} - \ldots = -\sum_{k=1}^{\infty}\frac{u^{2k}}{k}
\end{align*}
Based on the ratio test, the corresponding function is analytic for $|u| < 1$. Precisely,
\begin{align*}
\limsup_{k\to\infty}\left|\frac{a_{k+1}}{a_{k}}\right| = \limsup_{n\to\infty}\left(\frac{k}{k+1}\right)|u^{2}| = |u|^{2} < 1 \Longleftrightarrow |u| < 1
\end{align*}
Hence we can differentiate the obtained expression in order to get:
\begin{align*}
-\frac{2u}{1-u^{2}} = -2\sum_{k=1}^{\infty}u^{2k-1} \Rightarrow f(x) = \frac{x}{1-x^{2}} = \sum_{k=1}^{\infty}x^{2k-1} = x + x^{3} + x^{5} + \ldots
\end{align*}
whenever $|x| < 1$, and we are done.
Hopefully this helps!
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