Find the sum of power series $\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$ Find the sum of power series $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$$
How can we use derivation/integration method for this series?
 A: For $x\ge0,$ 
$$\dfrac{(-1)^nx^{n-1}}{2n-1}=\dfrac ix\dfrac{(i\sqrt x)^{2n-1}}{2n-1}$$
Now $$\ln(1+y)-\ln(1-y)=\sum_{r=1}^\infty\dfrac{y^{2r-1}}{2r-1}$$
A: $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$$
is convergent for $-1<x<1$. Then, in case of $0<x<1$ and with $X=x^{1/2}$ Let:
$$y(X)=\sum_{n=1}^{\infty}\frac{(-1)^nX^{2n-1}}{2n-1}$$
(wich is different from the above series. Do not confuse them. We will come back to the above series at the end).
$$y'(X)=\sum_{n=1}^{\infty}(-1)^nX^{2n-2}=-\frac{1}{1+X^2} \qquad \text{geometric series}$$
$$Y(X)=-\int \frac{dx}{1+X^2}=-\tan^{-1}(X)+c$$
$X=0 \quad\to\quad Y(0)=\sum_{n=1}^{\infty}\frac{(-1)^n 0^{2n-1}}{2n-1}=0=-\tan^{-1}(0)+c \quad\to\quad c=0$
$$y(X)=\sum_{n=1}^{\infty}\frac{(-1)^nX^{2n-1}}{2n-1}=-\tan^{-1}(X)$$
With $X=x^{1/2}$
$$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1/2}}{2n-1}=-\tan^{-1}(x^{1/2})$$
$$x^{-1/2}\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{2n-1}=\tan^{-1}(x^{1/2})$$
$$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{2n-1}=-x^{1/2}\tan^{-1}(x^{1/2})$$
Do similar calculus in case of $-1<x<0$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{n = 1}^{\infty}{\pars{-1}^{n}\,x^{n - 1} \over 2n - 1} & =
\sum_{n = 1}^{\infty}\pars{-1}^{n}x^{n - 1}\int_{0}^{1}y^{2n - 2}\,\dd y =
-\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-xy^{2}}^{n - 1}\,\dd y =
-\int_{0}^{1}{\dd y \over 1 + xy^{2}}
\\[5mm] & =
-\,{1 \over \root{x}}\int_{0}^{\root{x}}{\dd y \over y^{2} + 1} =
\bbx{\ds{-\,{\arctan\pars{\root{x}} \over \root{x}}}}
\end{align}
