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What is the power series expansion of this expression, and what is its radius of convergence?

$$\frac {(1+x)\sqrt[3] {1-3x^2}} {\sqrt {1+x^2}} $$

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    I want the steps not just the answer. And anyways, you plugged in $ \sqrt {1-3x^2} $ into wolfram, not $ \sqrt[3] {1-3x^2} $ – 2017-01-13
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    it was only a comment – 2017-01-13

1 Answers 1

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Hint: $$f(x)=\frac {(1+x)\sqrt[3] {1-3x^2}} {\sqrt {1+x^2}}=(1+x)(1-3x^2)^{1/3}(1+x^2)^{-1/2}.$$ $$(1+t)^p=\binom{p}{0}+\binom{p}{1}t+\binom{p}{2}t^2+\cdots=\sum_{k=0}^{+\infty}\binom{p}{k}t^k\quad (p\in\mathbb{R},\;|t|<1).$$

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    But how do I combine the expansions of $ \sqrt[3] {1-3x^2} $ and $ \frac {1}{\sqrt {1+x^2}} $ ? Can you please solve it in full if you have the time? – 2017-01-13
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    Use https://en.wikipedia.org/wiki/Cauchy_product. Try it, and we review your work. – 2017-01-13
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    Ok. I'll try it. Thank you very much. – 2017-01-13
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    Ok I got the sum alright by multiplying all the series together, which turns out to be:$ 1+x-\frac {3x^2}{2}-\frac {3x^3}{2}-\frac {x^4}{8}- $... But how do I find the interval in which it converges? – 2017-01-13
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    Right. Now $1+x$ coverges for all $x$, $(1-3x^2)^{1/3}$ iff $|3x^2|<1$ (that is iff $|x|<1/\sqrt{3}$) and $(1+x^2)^{-1/2}$ iff $|x^2|<1$ (that is iff $|x|<1$). So, $R=1/\sqrt{3}.$ – 2017-01-13
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    Thank you very much. I understand now :) – 2017-01-13