<s>[INST] Series expansion of $\frac{x^2}{1+ \sin x}$ For the series expansion at $x=0$ for
$\dfrac{x^2}{1+ \sin x}$ WolframAlpha gives $$x^2 -x^3 +x^4-\frac{5x^5}{6}+\frac{2x^6}{3}-\frac{61x^7}{120}+O(x^8)$$
But I'm missing something in the simplification, and I can't quite see it. How do I move from 
$\dfrac{x^2}{1+(x-\frac{x^3}{6}+O(x^5))}$ to that final form?
 [/INST] Note that $\frac{1}{1+x} = 1 - x + x^2 - ...$ as $x \rightarrow 0$. So, since $\sin(x) \rightarrow 0$ as $x \rightarrow 0$, we can write it as follows:
$\frac{x^2}{1+\sin(x)} = x^2\cdot (1 - \sin(x) + \sin^2(x) - \sin^3(x) + ...) \\
= x^2\cdot (1 - x + x^3/6 + x^2 - x^3 + \mathcal{O}(x^4)) \\
= x^2 - x^3 + x^4 - 5x^5/6 + \mathcal{O}(x^6).$
If more terms are desired, expand as required.
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