Find the sum of $\sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$ Find the sum of $S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1}$.
I know convergence radius is $1$ because $\frac{1}{\sqrt[n] {(2n-1)}} = \frac{1}{1} = 1.$
Then:
$$x^{-1} S(x) = \sum_{n=1}^\infty (-1)^{n+1}(2n-1)x^{2n+2}. $$
$$\int x^{-1}S(x) = \sum_{n=1}^{\infty} (-1)^{n+1}(2n-1) \frac{x^{2n-1}}{2n-1} = $$
$$ \sum_{n=1}^\infty (-1)^{n+1}x^{2n+1} = $$
$$ -\frac{x}{1-x} = \frac{x}{-1+x}. $$
Now getting back our original series:
$$x^{-1} S(x) = (\frac{x}{-1+x})' = \frac{(-1+x)-x}{(-1+x)^2} = \frac{-1}{1-2x+x^2}$$
$$xx^{-1}S(x) = S(x) = \sum_{n=1}^\infty (-1)^{n+1} (2n-1)x^{2n-1} = \frac{-x}{1-2x+x^2} = \frac{1}{-\frac{1}{x} + 2 - x}.$$
What is my problem?
wolframalpha.com  says the answer is wrong (it doesn't specify sum of series with $x$).
Thanks in advance!

$$S=\sum_{n=0}^\infty (-1)^n (2n+1)x^{2n+1}$$
Thus, $$\begin{align}
S&=\sum_{n=0}^\infty (-1)^n(2n)x^{2n+1}+\sum_{n=0}^\infty (-1)^nx^{2n+1}\\
&=2x\sum_{n=0}^{\infty}n(-x^2)^{n}+\frac{x}{1+x^2}\\
&=2x\left(\sum_{n=0}^\infty(n+1)(-x^2)^n-\sum_{n=0}^\infty (-x^2)^n\right)+\frac{x}{1+x^2}\\
&=2x\left(\frac{1}{(1+x^2)^2}-\frac{1}{1+x^2}\right)+\frac{x}{1+x^2}\\
&=\frac{2x}{(1+x^2)^2}-\frac{x}{1+x^2}=\frac{x-x^3}{(1+x^2)^2}
\end{align}$$
Well, that's my way.  In reference to your calculus,
$$\begin{align}
S&=x\frac{\mathrm d}{\mathrm dx}\left(\sum_{n=0}^\infty (-1)^nx^{2n+1}\right) \\
&=x\frac{\mathrm d}{\mathrm dx}\left(x\sum_{n=0}^\infty(-x^2)^n\right) \\
&=x\frac{\mathrm d}{\mathrm dx}\left(\frac{x}{1+x^2}\right) \\
&=x\left(\frac{1(1+x^2)-x(2x)}{(1+x^2)^2}\right)=\frac{x(1-x^2)}{(1+x^2)^2}
\end{align}$$
