Big $\mathcal{O}$ notation

The estimate $$ \frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}3+\frac{x^5}5+O\left(x^7\right)\tag{1} $$ is only for small $x$. For $\left|x\right|\le a$, we have that $\left|x\right|^7\le a\left|x\right|^6$; therefore, $(1)$ implies $$ \frac12\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}3+\frac{x^5}5+O\left(x^6\right)\tag{2} $$ However, $(1)$ gives more information (that is, a closer approximation for small $x$) than $(2)$.