Expand ln (1+x)/(1-x) in a power series in x

Expand $\ln\dfrac{1+x}{1-x}$ in a power series in $x$.

Explanation of Method

Splitting the log of a quotient into $\ln(1+x)-\ln(1-x)$ and subtracting the two series causes every even-power term to cancel exactly, leaving a series with only odd powers and doubled coefficients — the series for $2\operatorname{artanh}(x)$.

Split $\ln\frac{1+x}{1-x}=\ln(1+x)-\ln(1-x)$.
Subtract the two Maclaurin series and observe the cancellation.
Determine the region of validity.

Step 1 · Split the logarithm

$$ \ln\frac{1+x}{1-x} = \ln(1+x)-\ln(1-x). $$

Step 2 · Expand and subtract

$$ \ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}nx^n, \qquad \ln(1-x)=-\sum_{n=1}^{\infty}\frac1nx^n. $$ Subtracting: for even $n$, the terms cancel; for odd $n$, they double. Explicitly, $$ \ln\frac{1+x}{1-x} = 2x+\frac{2x^3}3+\frac{2x^5}5+\cdots = 2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}. $$

Step 3 · Region of validity

$\ln(1+x)$ converges on $-1

Summary Table

Property	Value
Series	$2\sum x^{2n+1}/(2n+1)$
Closed form	$2\operatorname{artanh}(x)$
Region of validity	$\|x\|<1$

Additional Notes

The even-power cancellation is a direct consequence of the odd symmetry of $\ln\frac{1+x}{1-x}$ under $x\mapsto-x$: swapping $x$ and $-x$ negates the whole expression, so only odd powers can survive.
This is exactly $2\operatorname{artanh}(x)$, the inverse hyperbolic tangent — the direct analogue of $\arctan x$ (Problem 267) with all signs made positive, reflecting $\operatorname{artanh}(x)=-i\arctan(ix)$.
Unlike $\arctan x$'s series, which converges (conditionally) at both endpoints, this series diverges at both $x=\pm1$ since the underlying function itself has genuine singularities there ($\ln\frac20$ and $\ln\frac02$).

Result

$$ \ln\frac{1+x}{1-x} = 2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}, \quad |x|<1. $$