4
The bug is fixed at least since v12.2:
Series[Pochhammer[1 + n, n], {n, Infinity, 1}]
(* E^SeriesData[n, Infinity, {-1 + 2*Log[2] + Log[n]}, -1, 3, 1]*
SeriesData[n, Infinity, {Sqrt[2], 0, -1/12*1/Sqrt[2]}, 0, 4, 2] *)
Plot[{(2^(2 n) Gamma[1/2 + n])/Sqrt[π], % // Normal} // Evaluate, {n, 0, 2},
PlotStyle -> {Automatic, Dashed}]
3
$Version
(* "12.3.0 for Mac OS X x86 (64-bit) (May 10, 2021)" *)
Clear["Global`*"]
x << 1 and x >> 1 represent Asymptotic (v12.1 or later) behavior.
For x << 1
Asymptotic[f[x], {x, 0, 4}]
(* 1 - x - x^3 *)
In earlier versions
Series[f[x], {x, 0, 4}] // Normal
(* 1 - x - x^3 *)
For x >> 1
Asymptotic[f[x], {x, ...
3
Maybe this post answers your question: https://math.stackexchange.com/questions/51770/taylor-expansion-at-infinity
Expansion at large x is equivalent to expansion at x-->1/x for small x.
f[x_] := ((1 + x (1 - Sqrt[1 + x^2]))^2 - x +
x^3 (1 - Sqrt[1 + x^2])^2)/(1 + x^2 (1 - Sqrt[1 + x^2])^2)
Normal[Series[f[x], {x, Infinity, 4}]]
Normal[Series[f[1/x]...
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