The Peano form of the remainder of $\frac1{1+x^2}$'s Maclaurin formula? I need to get the Maclaurin formula with the Peano form of the remainder of $\displaystyle \frac{1}{1+x^2}$. Here is what I do:
Since we have $$\displaystyle \frac{1}{1+x}=1-x+x^2+...+(-1)^nx^n+o(x^n)$$
then $$\displaystyle \frac{1}{1+x^2}=1-x^2+x^4+...+(-1)^nx^{2n}+o(x^{2n})$$
But in my textbook the remainder is different:
$$\displaystyle \frac{1}{1+x^2}=1-x^2+x^4+...+(-1)^nx^{2n}+o(x^{2n+1})$$
I can't figure out why the remainder can be $o(x^{2n+1})$, which is stronger than $o(x^{2n})$. Could anyone help me out here?

The function $$x \mapsto\frac1{1+x^2},\qquad x \in (-1,1),$$is even, then there is no term in $x^{2n+1}$ and you are allowed to write $o(x^{2n})=o(x^{2n+1})$.
