I think I have found two massive errors on Math.com but I throw my logic out here for a third party to verify as I've been doing Contour Integrals for about 10 hours straight now so I am very tired.
If you navigate to Math.com's logarithmic expansion page the following identities are given at the bottom:
$$ \tag{1} \sum_{n=1}^{\infty}\frac{(-1)^{n}x^n}{n} = -x +\frac{1}{2}x^2 - \frac{1}{3}x^3 + \frac{1}{2}x^4 + \cdots = -\ln{(x)} \quad\text{if}\ -1 < x \leq 1 $$
and $$ \tag{2} \sum_{n=1}^{\infty}\frac{x^n}{n} = x +\frac{1}{2}x^2 + \frac{1}{3}x^3 + \frac{1}{2}x^4 + \cdots = -\ln{(x+1)} \quad\text{if}\ -1 < x \leq 1 $$
However, the Mercator series is well known and defined as,
$$ \ln{(x+1)} = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{2}x^4 + \cdots \quad\text{if}\ -1 < x \leq 1 $$
Taking the known result for the Mercator series, we can can show that Equation 1, is clearly incorrect, $$ \tag{3} \ln{(x+1)} = (-1)\sum_{n=1}^{\infty}\frac{(-1)^{n}x^n}{n} \ne (-1)\ln{(x)} $$ as this is the negative of Math.com's definition in Equation 1!
Multiplying both sides of Equation 3 by $-1$, we get, $$ -\ln{(x+1)} = \sum_{n=1}^{\infty}\frac{(-1)^{n}x^n}{n} \ne \sum_{n=1}^{\infty}\frac{x^n}{n} $$ but this clearly is not equal to the second identity given on Math.com and they are missing a factor of $(-1)^n$ in the summation.
The Mercator series is indeed $$ \log(1+x)=\sum_{n\ge1}\frac{(-1)^{n+1}x^n}{n}= -\sum_{n\ge1}\frac{(-1)^{n}x^n}{n} $$ so the series (1) is just $-\log(1+x)$ (convergent for $-1<x\le 1$).
If we change $x$ into $-x$, we get $$ -\log(1-x)=\sum_{n\ge1}\frac{(-1)^n(-x)^n}{n}= \sum_{n\ge1}\frac{x^n}{n} $$ which converges for $-1\le x<1$.
There is clearly a typo in relation (1) (missing $1\:+$); relation (2) is clearly wrong for $x=1/2$, because the left-hand side is positive and the right-hand side is negative. With $1-x$, instead, it's correct.