I defined the function G[n] by inputting:
G[n_] := Product[2^n - 2^i, {i, 0, n - 1}]
I want to know the limit as n goes to infinity of G[n]/2^(n^2) so I input:
Limit[G[n]/2^(n^2), n -> ∞]
Mathematica returns 1, which I don't think is true.
$\begingroup$
$\endgroup$
4
asked Jun 11, 2017 at 12:47
1 Answer
$\begingroup$
$\endgroup$
1
First note that the sequence $G_n:=\prod_{i=0}^{n-1} (2^n-2^i)$ is A002884 in OEIS. Writing $$G_n= 2^{n(n-1)/2} \prod_{i=1}^n (2^i-1) $$ we see that $$ H_n:=2^{-n^2/2} G_n =2^{-n(n+1)/2} \prod_{i=1}^n (2^i-1) $$ Mathematica verifies that $$ H_n=\prod_{i=1}^n(1-2^{-i})$$ Thus $$ \lim_{n\rightarrow\infty} H_n = \prod_{n=1}^\infty (1-2^{-i})=0.28878\,80950\ldots$$
-
1$\begingroup$ Just to add to that, the closed form for the limit is
In[40]:= Product[1 - 2^-i, {i, 1, ∞}] Out[40]= QPochhammer[1/2, 1/2] In[41]:= N[%] Out[41]= 0.2887880950866024$\endgroup$ Commented Jun 12, 2017 at 16:38
f[n_] = Simplify[G[n]/2^(n^2), n>0]which simplifies toQPochhammer[2^(-n), 2, n]andLimit[f[n], n->Infinity] == QPochhammer[0, 2, Infinity] == QPochhammer[0, 2] == 1evaluates toTrue. Trying to verify this numerically gets unbearably slow for very largen. $\endgroup$QPochhammer[2^(-n), 2, n] /. n -> 1000.0gives1.0, whereasQPochhammer[2^-n, 2, n] /. n -> 1000 // Ngives0.288788. $\endgroup$