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The series expansion of $\displaystyle\left(1-\frac{a}{x}\right)^{1/3}$ using the Binomial theorem is given by $$\displaystyle\left(1-\frac{a}{x}\right)^{1/3}=1-\frac{a}{3x}-\frac{a^{2}}{9x^{2}}-\ldots$$

How may I employ Mathematica to obtain the same result?

It should be remarked here that I previously read other posts and tried to compute the expansion using the command Series. However, I am not obtaining the same result. That is to say if we employ the Series command,

Series[(1 - (a/x))^(1/3), {x, 0, 2}, {y, 0, 4}]

we shall have

$$\sqrt[3]{-\frac{a}{x}}-\frac{x \sqrt[3]{-\frac{a}{x}}}{3 a}-\frac{x^2 \sqrt[3]{-\frac{a}{x}}}{9 a^2}+O\left(x^3\right).$$

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  • $\begingroup$ Why did you write {y, 0, 4} when there is no $y$ in the expression you’re expanding? $\endgroup$
    – Ghoster
    Mar 16, 2023 at 4:27

2 Answers 2

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$Version

(* "13.2.1 for Mac OS X ARM (64-bit) (January 27, 2023)" *)

Clear["Global`*"]

f[x_] := (1 - a/x)^(1/3)

Do a series expansion about Infinity

Series[f[x], {x, Infinity, 3}]

enter image description here

The coefficient of x^-n is

c[n_] = Assuming[n >= 0,
  SeriesCoefficient[f[x], {x, Infinity, n}]]

(* (-a)^n Binomial[1/3, n] *)

Assuming[n ∈ NonNegativeIntegers, 
 c[n] // FunctionExpand // FullSimplify]

(* ((-a)^n Gamma[4/3])/(n! Gamma[4/3 - n]) *)

Verifying,

f[x] === (Sum[c[n]*x^-n, {n, 0, Infinity}] // Simplify)

(* True *)
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You are almost there. Try the following:

Series[(1 - (a/x))^(1/3), {a, 0, 2}] // Normal

(* 1 - a^2/(9 x^2) - a/(3 x) *)

Have fun!

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