7
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Bug introduced after 5.2, fixed between 12.0.0 and 12.1.1.


The function

  Pochhammer[1 + n, n]

tends to infinity. We have

  FunctionExpand[Pochhammer[1 + n, n]]
  (* (2^(2*n)*Gamma[1/2 + n])/Sqrt[Pi] *)

But following result

  Series[Pochhammer[1 + n, n] , {n, Infinity, 1}]
  (* 1 + O[1/n]^2 *)

is a bug! Tested under versions 10.2 and 7.0, same results. Interesting is that

  Limit[Pochhammer[1 + n, n]/(2^(2*n + 1/2)*n^n/E^n), n -> Infinity]

is correct.

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5
  • $\begingroup$ What is your question here? Have you reported this behavior to Wolfram support? $\endgroup$
    – MarcoB
    Commented Apr 24, 2016 at 4:50
  • $\begingroup$ About bugs please see mathematica.stackexchange.com/questions/84077/… The bug was fixed only after I put the question here... $\endgroup$ Commented Apr 24, 2016 at 6:06
  • $\begingroup$ Question is: please confirm that it is a bug and test it in other versions. $\endgroup$ Commented Apr 24, 2016 at 6:10
  • 1
    $\begingroup$ Of course, the workaround is to use Series[] on the gamma function expression instead. Anyway, 5.2 does not have the bug, but 10.4.1 does. $\endgroup$ Commented Apr 24, 2016 at 13:34
  • $\begingroup$ I wonder if MMA is actually being smart, figuring out that there is no Taylor Series at infinity, and instead giving the MacClaurin series for the singuarity "at infinity". Note that for simple poles, (e.g., Series[1/Sin[x], {x, 2 Pi, 5}]), it does find the negative power, though it doesn't include anything like O[1/x]^2. ...just a thought. $\endgroup$
    – Paco Jain
    Commented Apr 25, 2016 at 19:41

1 Answer 1

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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}]

Mathematica graphics

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2
  • $\begingroup$ I can confirm that it is fixed in 12.1.1 but broken in 12.0.0 (MacOS X 10.14) $\endgroup$
    – QuantumDot
    Commented Jun 13, 2021 at 0:50
  • $\begingroup$ @QuantumDot Thx for checking! You can directly edit, this is a community wiki :) . $\endgroup$
    – xzczd
    Commented Jun 13, 2021 at 9:24

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