<s>[INST] How is $x^2+1=(1/{x^2})[1-{1}/{x^2}+{1}/{x^4}-{1}/{x^6}+\cdots]$? The author of my book writes:
$$x^2+1=x^2\left(1+\frac{1}{x^2}\right)$$
$$=\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]$$
I do not understand the last step. How did the author write the last step. Please help.
 [/INST] I don't think they are equal. As others have said
$$
\frac{x^2}{1+x^2} = 1 - \frac{1}{x^2} + \frac{1}{x^4} \dots \text{ for } \left\lvert \frac{-1}{x^2} \right\rvert < 1
$$
so for $x = \sqrt{2}$
$$
\frac{1}{x^2} \left( 1 - \frac{1}{x^2} + \frac{1}{x^4} \dots \right) = \frac{1}{2} \left( \frac{2}{3} \right) = \frac{1}{3}.
$$
However $x^2+1 = 3 \neq \frac{1}{3}$.
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