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I want to evaluate the following Series:

Series[(x^a + x)^b, {x, ∞, 1}, Assumptions :> {a > 1, b > 1}], 

but Mathematica simply returns backs the original expression:

(x + x^a)^b

What should I do to get the Series?

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  • $\begingroup$ If the expression is divergent when x -> Infinity, there is no guarantee that a series expansion exists at infinity. Are you sure that the question you're asking even has an answer? The fact that Mathematica returns the expression unchanged hints (though by no means is proof) that this may not be the case. Furthermore, Assumptions should be specified as Assumptions -> (a > 1 && b > 1), though it does not affect Mathematica's answer in this case. $\endgroup$ Commented Jun 5, 2017 at 9:16

2 Answers 2

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To get Mathematica to factor and manipulate expressions with variables in the exponent can seem difficult. In particular Series won't expand an expression like the OP's. Here's an approach that converts the expression into x^(a*b) times an expression that can be expanded in terms of u == x^(1-a). This is done under the assumptions that a, b, x are positive, and further that a > 1 so that the power 1-a may be assumed to be negative.

With[{factor = x^(a*b)},
 expr = factor * Simplify[((x^a + x)^b/factor)^(1/b), b > 0 && a > 0 && x > 0]^b
 ]
(*  x^(a b) (1 + x^(1 - a))^b  *)

sub = x^(1 - a) -> u;
Series[expr /. sub, {u, 0, 3}]
Normal@% /. Reverse[sub]

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The main problem is with the $x^a+x$ term. You could try substituting $x\to \frac{1}{x}$ and use explicit values of a to try to discern a pattern:

Series[(1/x^3 + 1/x)^b, {x, 0, 8}] //TeXForm

$\left(\frac{1}{x^3}\right)^b \left(1+b x^2+\frac{1}{2} (b-1) b x^4+\frac{1}{6} (b-2) (b-1) b x^6+\frac{1}{24} (b-3) (b-2) (b-1) b x^8+O\left(x^9\right)\right)$

Series[(1/x^4 + 1/x)^b, {x, 0, 12}] //TeXForm

$\left(\frac{1}{x^4}\right)^b \left(1+b x^3+\frac{1}{2} (b-1) b x^6+\frac{1}{6} (b-2) (b-1) b x^9+\frac{1}{24} (b-3) (b-2) (b-1) b x^{12}+O\left(x^{13}\right)\right)$

Series[(1/x^5 + 1/x)^b, {x, 0, 16}] //TeXForm

$\left(\frac{1}{x^5}\right)^b \left(1+b x^4+\frac{1}{2} (b-1) b x^8+\frac{1}{6} (b-2) (b-1) b x^{12}+\frac{1}{24} (b-3) (b-2) (b-1) b x^{16}+O\left(x^{17}\right)\right)$

Inverting back, it looks like the pattern is:

ser[max_] := x^(a b) Sum[Binomial[b,n] x^(n(1-a)), {n, 0, max}]

We could test this. For example:

exact[x_,a_,b_]:=(x^a+x)^b
approx[x_,a_,b_]:=Evaluate[ser[20]]

exact[1`100*^10,2,3]
exact[1`100*^10,5,6]

1.000000000300000000030000000001000000000000000000000000000000000000000000000000000000000000000000000*10^60

1.00000000000000000000000000000000000000060000000000000000000000000000000000000015000000000000000000*10^300

approx[1`100*^10,2,3]
approx[1`100*^10,5,6]

1.000000000300000000030000000001000000000000000000000000000000000000000000000000000000000000000000000*10^60

1.00000000000000000000000000000000000000060000000000000000000000000000000000000015000000000000000000*10^300

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