Show that $\displaystyle\lim_{n \to \infty} \prod_{k=1}^{n} \left(1-2^{k-1-n}\right)$ exists and strictly positive.

Question

I have a question regarding limit calculations. As part of something I'm working on, I faced the challenge of calculating the limit $$ \lim_{n \to \infty}\prod_{k = 1}^{n} \left(1 - 2^{k-1-n}\right) $$ This was challenging for me and I couldn't solve this by now, although numerical calculation shows the limit exists and is strictly positive $\left(\approx 0.289\right)$.

I couldn't prove it. For my purposes it is sufficient to show the limit exists and is strictly positive. I will appreciate any help.

Answer 1

The problem can be solved elementary, giving an explicit lower estimate. No estimates for the $\log$ functions are needed. Observe that the expression is equal $$ \prod_{k=1}^n\left (1-2^{-k}\right )$$ We will make use of the generalized Bernoulli inequality $$(1+x_1)(1+x_2)\ldots (1+x_n)> 1+x_1+x_2+\ldots +x_n$$ where $x_i>-1$ and all $x_i$ are nonzero and have the same sign. The inequality can be proved easily by induction (cf. generalization of the base)

Thus $$\prod_{k=1}^n\left (1-2^{-k}\right )={1\over 2}\prod_{k=2}^n\left (1-2^{-k}\right )> {1\over 2}\left [1-\sum_{k=2}^n2^{-k}\right ]\\ > {1\over 2}\left [1-\sum_{k=2}^\infty2^{-k}\right ]={1\over 4}$$

Answer 2

The comments helped me to fully answer this question. We can change indices to see that:

$$\prod_{k=1}^{n}(1-2^{1-k-n}) = \prod_{k=1}^{n}(1-2^{-k}).$$

Note that the above sequence converges to a strictly positive limit iff the following series converges:

$$\sum_{k=1}^{\infty}\ln(1-2^{-k})$$

Since this is a non-positive sequence, we can use the comparison test with $-2^{-k}$ and get that this series indeed converges.

Answer 3

For the existence we can observe that

$$0\le \left|\log\left(\prod_{k=1}^{n} (1-2^{k-1-n})\right)\right|=-\sum_{k=1}^{n} \log\left(1-2^{k-1-n}\right) $$ $$\le 2\sum_{k=1}^{n}2^{k-1-n}=\frac1{2^{n-1}}\sum_{k=0}^{n-1}2^{k}=2-\frac{1}{2^{n-1}}\to 2$$

Answer 4

If you are familiar with the q-Pchhammer symbols $$P_n=\prod_{k=1}^{n} (1-2^{k-1-n})=\frac{2^{n+1}}{2^{n+1}-1}\,\,\left(2^{-(n+1)};2\right){}_{n+1}$$

The numerator form sequence $A005329$ in $OEIS$ (with a shift of $1$ and the denominator is $ 2^{\frac{ n (n+1)}{2}}$.

This meeans that $$P_n=2^{\frac{- n (n+1)}{2}}\,\prod_{k=1}^{n} (2^k-1)=\prod_{k=1}^{n}\left(1-2^{-k}\right)=\left(\frac{1}{2};\frac{1}{2}\right)_n$$

For the convergence $$\frac{P_{n+1}}{P_n}=1-2^{-(n+1)}$$

I did not find a simple asymptotic of $P_n$ without special functions, but it converges very fast $$\left( \begin{array}{cc} n & P_n \\ 10 & 0.2890702984197489333606512 \\ 20 & 0.2887883704965666901812882 \\ 30 & 0.2887880953555572936802431 \\ 40 & 0.2887880950868650725213162 \\ 50 & 0.2887880950866026777742536 \\ 60 & 0.2887880950866024215293835 \\ 70 & 0.2887880950866024212791443 \\ 80 & 0.2887880950866024212789000 \\ 90 & 0.2887880950866024212788997 \\ 100 & 0.2887880950866024212788997 \\ \end{array} \right)$$

and $$\left(\frac{1}{2};\frac{1}{2}\right)_{\infty }=0.2887880950866024212788997$$

The number $0.288788\cdots$ is not recognized by the $ISC$ which however proposed, as an upper bound, $$\log(\sqrt 3)\,\, \sqrt{\frac{5-\sqrt{5}}{10}} $$ which is an absolute error of $7.649\times 10^{-7}$

Edit

If we take logarithms, since $$\int \log(1-2^{-k})\,dk=\frac{\text{Li}_2\left(2^{-k}\right)}{\log (2)}$$ we could think about Euler-MacLaurin summation. Using its simplest form $$\log(P_n)= \frac{1}{2} \log \left(1-2^{-n}\right)+\frac{1+2^{-(n+2)}}{2^n\,\log (2)}+$$ $$\log(2)\,\sum_{m=1}^7 \frac {A_m}{2^{m n}\,(1-2^{-n})^m}+C$$ Using $L=\log(2)$, the coefficients are $$\left( \begin{array}{cc} m & A_m \\ 1 & \frac{-L^6+40 L^4-1680 L^2+100800}{1209600} \\ 2 & -\frac{21 L^6-200 L^4+1680L^2}{403200} \\ 3 & -\frac{301 L^6-1000 L^4+1680L^2}{604800} \\ 4 & \frac{8 L^4-7 L^6}{4032} \\ 5 & \frac{2 L^4-7 L^6}{2520} \\ 6 & -\frac{L^6}{480} \\ 7 & -\frac{L^6}{1680} \\ \end{array} \right)$$

For this level of expansion $$C=-\frac{-1561 L^8+1000 L^6-1680 L^4+16800 L^2+16800 \pi ^2}{201600 L}$$ and, for a infinite value of $n$, $P_\infty=e^C=0.2888769$.

Answer 5

$$\prod_{k=1}^\infty(1-x_k)=\sum_{k=0}^\infty(-1)^{k}e_k$$ where $e_k$ is the $k$-th elementary symmetric function of $x_1,x_2,x_3,...$

For $x_k=\frac1{2^k}$, we have that $$e_0=e_1=1>e_2=\frac13>e_3=\frac1{21}>...>e_k=\prod_{i=2}^k\frac1{2^i-1}>...$$ decreasing to zero. So, by alternating series test $\prod_{k=1}^\infty(1-\tfrac1{2^k})$ is convergent.

Note: The first ten terms of the series give $$\approx 0.28878809508660224213$$

Answer 6

Theorem: If every $x_j\in [0,1)$ then $\prod_{j=1}^{\infty}(1-x_j) >0$ iff $\sum_{j=1}^{\infty}x_j<\infty$.

Let $P(n)=\prod_{k = 1}^{n} \left(1 - 2^{k-1-n}\right)$. Then $P(n)=\prod_{j = 1}^{n} \left(1 - 2^{-j}\right)$.

Now $L=\lim_{n\to\infty}P(n)$ exists because $0<P(n+1)<P(n)$ for every $n$.

And $L>0$ by the above Theorem, with $x_j=2^{-j}$ and $L=\prod_{j=1}^{\infty}(1-2^{-j})$.

There is a companion theorem: If every $x_j\ge 0$ then $\prod_{j=1}^{\infty}(1+x_j)$ $<\infty$ iff $\sum_{j=1}^{\infty}x_j<\infty$.