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I know this product converge $$\prod_{k=1}^\infty (1-1/2^k),$$ but I don't know how to prove that this limite is different to zero. This is equivalent to prove that $$\sum^{\infty}_{k=1} \log(1-1/2^k)$$ converge. I can't get it. Could you give me any hint? Thank! I'm trying to prove that $$\frac{|GL_n(\mathbb{F}_2)|}{2^{n^2}} \to \alpha,$$ where $\alpha > 0$

John Kevin
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  • $$ \Pi_{k=0}^\infty $$ $$ \prod_{k=0}^\infty $$ The first line above is coded as \Pi_{k=0}^\infty and the second as \prod_{k=1}^\infty. That is why the second is standard usage, and I edited accordingly. – Michael Hardy Feb 02 '20 at 20:46
  • We usually don't say that an infinite product "converges" unless the limit is non-zero. Also you want $k \ge 1$ otherwise you're multiplying by $0$. – Trevor Gunn Feb 02 '20 at 21:01
  • If your product starts from index $k=0$, then $(1-1/2^0)=0$, so all the partial products are $0$, hence the infinite product is also $0$. Is this what you want? – bjorn93 Feb 02 '20 at 21:38
  • Not. Actually, the product starts in k = 1. I'm going to change it. – John Kevin Feb 03 '20 at 01:01

3 Answers3

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for the lower bound, the general result is that for $\delta_k \in (0,1)$
$0\leq \prod_{k=1}^\infty (1-\delta_k)$
and the inequality is strict iff $\sum_{k=1}^\infty \delta_k \lt \infty$

the easy direction
comes from using $1 + x \leq e^x$

the harder direction
(your direction) comes from using if $\sum_{k=1}^\infty \delta_k \lt \infty$ then for every $\epsilon \gt 0$ there is some $K$ such that $\sum_{k=K}^\infty \delta_k \lt \epsilon$. You know this is true e.g. by looking at the partial sums $s_n = \sum_{k=1}^n \delta_k$ and seeing that they are Cauchy. Select $\epsilon' := \frac{1}{3}$, $K'$ and you can bound the tail of your product as

$0\lt \frac{2}{3} = 1 -\frac{1}{3}\leq \prod_{k=K'}^\infty (1-\delta_k)$
This is a slightly generalized Bernouli inequality. This can be proven directly with induction, but the nice approach is to recognize it follows from the union bound in probability i.e.
$1-P\Big(\bigcup_{k=K'}^\infty A_{k}\Big) = P\Big(\bigcap_{k=K'}^\infty A_{k}^C\Big) = \prod_{k=K'}^\infty (1-\delta_k)$ with independent events $A_k$

so the tail is $\in (0,1)$ and necessarily $\big(\prod_{k=1}^{K'-1}(1-\delta_k)\big) \in (0,1)$ and so their product is $\in (0,1)$

user8675309
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If $x$ is positive and near $0$ then $$ -2x \le \log(1-x) \le -x. $$ This can be seen by observing that the function $x\mapsto\log(1-x)$

  • is equal to $0$ at $0$, and
  • has slope $-1$ at $0$, and
  • has a negative second derivative at $0,$ so the graph is concave downward.
Michael Hardy
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$\log(1-x)$ is a concave function on $[0,1)$, so for any $x\in[0,1/2]$ we have $$ \log(1-x)\geq -2\log(2) x $$ immediately implying $$ \sum_{k\geq 1}\log\left(1-\frac{1}{2^k}\right) \geq -2\log(2)\sum_{k\geq 1}\frac{1}{2^k} = -2\log(2) $$ and $$ \prod_{k\geq 1}\left(1-\frac{1}{2^k}\right) \geq \frac{1}{4}.$$ Much better approximations can be derived through the Mellin transform, as done by Marko Riedel here.
Creative telescoping is also a chance. Numerically the LHS is $\approx 0.288788095$.

Jack D'Aurizio
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