TITLE: Sum of $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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:
$$
\sum_{n=1}^{\infty}{{(-1)}^{n+1}\frac{{x}^{2n-1}}{2n-1}}=\arctan{x}
$$