In the book Short Calculus The author, in order to prove the taylor series for the arctangent, he presented the taylor expansion of $$\frac{1}{1+x^2}$$
Which is (the one of the author):$$\frac{1}{1+x^2} =1-x^2+x^4-x^6+…+(-1)^{n-1}x^{2n-2}+(-1)^{n}\frac{x^{2n}}{1+x^2}$$
But using long division, I’ve found a slightly different one:
$$\frac{1}{1+x^2} =1-x^2+x^4-x^6+…+(-1)^{n}x^{2n}+…$$
Which one is correct? This is an important thing because it will change the remainder term of the taylor expansion for the arctangent.
Best Answer
The expansion is based on the high-school identity: $$1-x^n=(1-x)(1+x+x^2+\dots+x^{n-1})$$ which can be rewritten as $$\frac 1{1-x}=1+x+x^2+\dots+x^{n-1}+\frac{x^n}{1-x}.$$ Now substitute $-x^2$ to $x$ in this formula, and you obtain exactly the formula in the book.
Other than that, you also can obtain the same result, not by long division, but by the division along increasing powers of $1$ by $1+x^2$. As an example, here is this division up to order $8$: \begin{array}{rcl} &&\phantom{-}1-x^2+x^4-x^6 \\ 1+x^2&\Bigl(&\phantom{-}\not1 \\ && -\not1-x^2 \\ && \phantom{-1}\enspace+x^2 +x^4\\ && \phantom{-1\enspace+x^2} -x^4- x^6\\ && \phantom{-1\enspace+x^2 -x^4}+ x^6+x^8\\ && \phantom{-1\enspace+x^2 -x^4+ x^6}+x^8\\ \end{array} So the remainder at order $8$ is $x^8$, and this division means that $$1=(1-x^2+x^4-x^6)(1+x^2)+x^8\implies\frac 1{1+x^2}=1-x^2+x^4-x^6+\frac{x^8}{1+x^2}.$$