Sum of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$ I've been working with the series:
$$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$$
From the ratio test it is clear that the series converges for $|x| < 1$, but I'm unable to obtain the sum of the series.
I'm looking for any hint of how to obtain the sum.
Thanks in advance!
 A: Let $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{2n-1}$.  Then we have
$$\begin{align}
f'(x)=\sum_{n=1}^{\infty} (-1)^{n+1}x^{2n-2}&=\frac{-1}{x^2}\sum_{n=1}^{\infty} (-x^2)^{n}\\\\
&=\frac{1}{1+x^2} \tag 1
\end{align}$$
Integrating $(1)$ and using $f(0)=0$ reveals that 
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\arctan(x)}$$
A: Let $y$ be our sum so:
$$
y=\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}
$$
Let's differentiate it to get:
$$
y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{x}^{2n-2}}\\
y'=\sum_{n=1}^{\infty}{{(-1)}^{n-1}{x}^{2n-2}}\\
y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{(-x^2)}^{n-1}}\\
y'=\frac{1}{1+x^2}
$$
Now integrate to get the original sum to get:
$$
y=\arctan{x} +C
$$
It is easy to see that for $x=0$ we have $y=0$, so $C=0$, hence the sum equals:
$$
\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}=\arctan{x}
$$
A: $$(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1}=i^{2(n+1)}\cdot\dfrac{x^{2n-1}}{2n-1}=-i\cdot\dfrac{(ix)^{2n-1}}{2n-1}$$
If $S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1},$
$$i S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{(ix)^{2n-1}}{2n-1}$$
Now for $-1<y\le1,\ln(1+y)=y-\dfrac{y^2}2+\dfrac{y^3}3-\dfrac{y^4}4+\cdots$
$\ln(1-y)=-y-\dfrac{y^2}2-\dfrac{y^3}3-\dfrac{y^4}4-\cdots$
$\ln(1+y)-\ln(1-y)=?$
$$\implies2i S=\ln(1+ix)-\ln(1-ix)=\ln\dfrac{1+ix}{1-ix}$$
Let $1=r\cos A,x=r\sin A, x=\tan A$
Now, $$\ln\dfrac{1+ix}{1-ix}=\ln(e^{2iA})=2iA=2i\arctan x$$
