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)?
 A: As an alternate without using $\ln x$. Make use of the geometric series:
$$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n \implies \frac{1}{1-x^2} = \sum_{n=0}^\infty x^{2n} \implies \frac{x}{1-x^2} = \sum_{n=0}^\infty x^{2n+1} $$
A: 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!
A: Hint:
$$\int{ \frac{x}{1-x^2} dx } = -\frac{1}{2} \ln{(1-x^2)}$$
A: Notice that:
$$\int f(x)dx = \int \dfrac{x}{1-x^2}dx=-\frac{1}{2}\int \frac{du}{u}=-\frac{1}{2}\ln|1-x^2|+C$$
(via the $u$-substitution $u=1-x^2$)
From your note:
$$\ln(x)=(x-1)-\frac{1}{2}(x-1)^2+\frac{1}{3}(x-1)^3+\dots$$
So substitute $1-x^2$ into $x$ in the natural log taylor expansion, and take the derivative to obtain a series representation of $f(x).$
A: 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.
$$\ln(x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n} \implies \ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x)^n}{n}$$
$$\frac{d}{dx}\ln(x+1) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{n(x)^{n-1}}{n} \implies \frac{1}{1+x} = \sum_{n=1}^{\infty} (-1)^{n+1}x^{n-1}$$
$$\frac{1}{1+(-x)} = \sum_{n=1}^{\infty} (-1)^{n+1}(-x)^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} (-1)^{n+1}(-1)^{n-1}x^{n-1} \implies \frac{1}{1-x} = \sum_{n=1}^{\infty} x^{n-1}$$
$$ \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2(n-1)} \implies \frac{1}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-2}$$
$$\frac{x}{1-x^2} = \sum_{n=1}^{\infty} x^{2n-1}$$
