# The limit of a product to infinity

What is the difficulty with this question to Mathematica? Is it just a problem with Mathematica v.8.0
or all versions are in trouble with it? Or is something wrong with my input?

Limit[Product[(1 + t/n^3)^(1/t^(1/3)), {n, 1, Infinity}],  t -> Infinity]

-
it's running for a long period of time is your problem ? –  Rorschach Jul 8 '13 at 18:18
It's running for a long period of time and gives no answer (1 hour is it OK?). The limit has a closed form though. –  Chris's sis Jul 8 '13 at 18:20
Instead of using infinity use a large number you shall get quite relevant result. –  Rorschach Jul 8 '13 at 18:27

I replaced t -> u^3, t^(1/3) -> u as a manual simplification. Then I got

Product[(1 + u^3/n^3)^(1/u), {n, 1, Infinity}]
(* (1/(Gamma[1 + u] Gamma[1 - u/2 - 1/2 I Sqrt[3] u] *
Gamma[1 - u/2 + 1/2 I Sqrt[3] u]))^(1/u) *)

Limit[
(1/(Gamma[1 + u] Gamma[1 - u/2 - 1/2 I Sqrt[3] u] *
Gamma[1 - u/2 + 1/2 I Sqrt[3] u]))^(1/u),
u -> Infinity]
(* E^((2 \[Pi])/Sqrt[3]) *)


Now Product returned a value for both forms of input (simplified and unsimplified). But the limit of the product (of the unsimplified input) takes forever. (Tested on V9.0.1 and V8.0.4.)

-
Hmmm... without using simplifications I obtained: (1/(Gamma[1 + t^(1/3)] Gamma[1/2 (2 + (-1 - I Sqrt[3])t^(1/3))] Gamma[ 1/2 (2 + I (I + Sqrt[3]) t^(1/3))]))^(1/t^(1/3)) –  Sektor Jul 8 '13 at 19:03
@NikolaDimitrov Yes, that's what I got. But finding the Limit of it takes forever. One can simplify before or after finding the Product. Edited to clarify (I hope). –  Michael E2 Jul 8 '13 at 19:41
@MichaelE2 thanks for your job (+1) –  Chris's sis Jul 8 '13 at 19:43