$$\ln(2)=\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{9}+\frac{1}{10}-\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+...$$ This is something I came up with and was intrigued, and no this isn't random; there's a pattern to this.
Since you guys are asking for a relation, here it is; $$\ln(2)=1+\sum_{n=1}^∞ \frac{(-1)^{S_n-1}}{S_n-1} $$
Where $S_n$ is the $n$-th number that is not a perfect power (A007916, and $S_1$ is $2$). Proof to this is also somewhat easy to derive.
Equivalently, the problem is to compute $L=\sum_{s\in S}(-1)^s/(s-1)$ where $S$ is the set of all perfect powers. A similar problem of computing $\sum_{s\in S}1/(s-1)\color{gray}{[{}=1]}$ goes back to Goldbach as far as I know.
The idea is pretty much the same. Let $N$ be the set of all integers (strictly) greater than $1$. Then each $s\in S$ has a representation $s=a^n$ with unique $a\in N\setminus S$ and $n\in N$; the same is true for $s\in N$ if we also allow $n=1$. Hence (each of the series below is absolutely convergent) \begin{align} L&=\sum_{a\in N\setminus S}(-1)^a\sum_{n=2}^{\infty}\frac{1}{a^n-1}\\&=\sum_{a\in N\setminus S}(-1)^a\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}a^{-nk}\\&=\sum_{a\in N\setminus S}(-1)^a\sum_{k=1}^{\infty}\frac{1}{a^k(a^k-1)}\\&=\sum_{s\in N}\frac{(-1)^s}{s(s-1)}=2\ln 2-1. \end{align} [The sum in the question is $1+\sum_{s\in N\setminus S}(-1)^{s-1}/(s-1)=1-(\ln 2-L)$.]