8
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Bug introduced in 9.0 or earlier and fixed in 11.0.0


This infinite product is correct:

Product[1/(1 - 1/2^n), {n, 1, Infinity}]
(* 1/QPochhammer[1/2, 1/2] *)

Numerical value of this result is not equal to zero

N[1/QPochhammer[1/2, 1/2]]
(* 3.462746619455064 *)

But from following equivalent expression I get a wrong output:

Product[1/(1 - 1/2^(n-1)), {n, 2, Infinity}]
(* 0 *)
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7
  • $\begingroup$ I cannot reproduce your final result, instead obtaining 1/(2 QPochhammer[1/2, 1/2]), as it should be. $\endgroup$
    – bbgodfrey
    Dec 19, 2015 at 17:18
  • $\begingroup$ I can confirm this using: 10.0 for Microsoft Windows (64-bit) with Windows 8.1 $\endgroup$
    – mattiav27
    Dec 19, 2015 at 17:21
  • $\begingroup$ Which version are you using? $\endgroup$
    – mattiav27
    Dec 19, 2015 at 17:22
  • 1
    $\begingroup$ I get 0/QPochhammer[2, 1/2] on 10.3.1. $\endgroup$ Dec 19, 2015 at 17:24
  • 1
    $\begingroup$ I get zero under 10.0.1 on OS X10.10.5, but if we change Infinity to some large integer instead, it evaluates correctly. $\endgroup$
    – march
    Dec 19, 2015 at 17:48

1 Answer 1

8
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$Version

"10.3.1 for Mac OS X x86 (64-bit) (December 9, 2015)"

p = Product[1/(1 - 1/2^n), {n, 1, Infinity}]

(*  1/QPochhammer[1/2, 1/2]  *)

Error:

Product[1/(1 - 1/2^(n - 1)), {n, 2, Infinity}]

(*  0/QPochhammer[2, 1/2]  *)

Workarounds:

Product[1/(1 - 1/2^(n - 1)), {n, m, Infinity}] /. m -> 2

(*  1/QPochhammer[1/2, 1/2]  *)

Limit[Product[1/(1 - 1/2^(n - 1)), {n, m, Infinity}], m -> 2]

(*  1/QPochhammer[1/2, 1/2]  *)

Product[1/(1 - 1/2^(n - 1)), {n, 2, Infinity}, Regularization -> "Dirichlet"]

(*  1/QPochhammer[1/2, 1/2]  *)
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