<s>[INST] Asymptotic behaviour of sequences Could anyone explain in details how these approximations as $n \to \infty$ are found? ($a$ is a positive real number)


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*${x_n} = \frac{1}{n}\left( {\frac{a}{3} - \frac{3}{2}} \right) + O\left( {\frac{1}{{{n^3}}}} \right)$ for ${x_n} = \sqrt[3]{{{n^3} + an}} - \sqrt {{n^2} + 3} $

*${y_n} = \frac{{{{( - 1)}^n}}}{{{n^{\frac{a}{2}}}}} - \frac{1}{{2{n^{\frac{{3a}}{2}}}}} + o\left( {\frac{1}{{{n^{\frac{{3a}}{2}}}}}} \right)$ for ${y_n} = \frac{{{{( - 1)}^n}}}{{\sqrt {{n^a} + {{( - 1)}^n}} }}$


The idea is to study the behaviour of a series whose $n-$th term is given by the above expressions in terms of the positive real $a$ and these asymptotic developments which I don't understand are part of the solution.
 [/INST] We have $\sqrt[3]{n^3+an}=n\sqrt[3]{1+\frac{a}{n^2}}$.  We expand the latter using Newton's Binomial Theorem to get $$\sqrt[3]{n^3+an}=n\sum_{k\ge 0} {1/3 \choose k} \left(\frac{a}{n^2}\right)^k=n(1+\frac{a}{3n^2}-\frac{a^2}{9n^4}+O(n^{-6}))$$
Repeating, we have $\sqrt{n^2+3}=n\sqrt{1+\frac{3}{n^2}}$, so 
$$\sqrt{n^2+3}=n\sum_{k\ge 0}{1/2 \choose k}\left(\frac{3}{n^2}\right)^k=n(1+\frac{3}{2n^2}-\frac{3^2}{8n^4}+O(n^{-6}))$$
Subtracting, we get $$\sqrt[3]{n^3+an}-\sqrt{n^2+3}=n\left(\frac{a}{3n^2}-\frac{3}{2n^2}-\frac{a^2}{9n^4}+\frac{9}{8n^4}+O(n^{-6})\right)$$
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