I'm learning about the big $\mathcal{O}$ notation and I'm a bit confused. Why is it that we can write things like $$\displaystyle x+\frac{x^3}{3}+\frac{x^5}{5}+\mathcal{O}(x^6)$$ when there's no $x^6$ term? Wouldn't it make sense to write $$\displaystyle x+\frac{x^3}{3}+\frac{x^5}{5}+\mathcal{O}(x^7)$$ instead? This is for the $\tanh^{-1}{x}$ series if it makes a difference.
Answer
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 $(2)$.
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