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it neither seems to be an expansion of some function nor in a telescopic form.

When $k=1$, $\prod_{i=1}^{\infty}1-\frac{1}{2^i} = 0.288....$ by Wolfram alpha, so the limit does exist but wolfram doesn't give a concrete value.

Any idea to get a precise value for this product?

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  • $\begingroup$ Wolfram Alpha definitely gives an answer - it gives $\left(2^{-k} ;\frac{1}{2}\right)_\infty$ $\endgroup$ – Brevan Ellefsen Nov 29 '16 at 2:31
  • $\begingroup$ This is a product, not as series btw. To your point, the way WA writes it is basically just another form of the product $\endgroup$ – Brevan Ellefsen Nov 29 '16 at 2:33
  • $\begingroup$ Given k >= 1, the limit does exist and is monotonically increasing in k and asymptotically approaching 1, but Wolfram doesn't show a closed form but resorts to q-Pochammer symbols, which is way beyond my capabilities. $\endgroup$ – Avraham Nov 29 '16 at 2:42
  • $\begingroup$ See here $\endgroup$ – user378947 Nov 29 '16 at 2:45
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I am not sure that this is an answer.

Using the q-Pochhammer symbol, the expression writes $$\prod_{i=k}^{\infty}1-\frac{1}{2^i}=\left(2^{-k};\frac{1}{2}\right){}_{\infty }$$ just as Brevan Ellefsen commented earlier.

I do not know any closed form of the result. Below is a table of the numerical values obtained $$\left( \begin{array}{cc} k & \left(2^{-k};\frac{1}{2}\right){}_{\infty }\\ 1 & 0.2887880950866024 \\ 2 & 0.5775761901732048 \\ 3 & 0.7701015868976065 \\ 4 & 0.8801160993115502 \\ 5 & 0.9387905059323200 \\ 6 & 0.9690740706398140 \\ 7 & 0.9844561987452080 \\ 8 & 0.9922078223573750 \\ 9 & 0.9960988334254430 \\ 10 & 0.9980481462110120 \end{array} \right)$$

Inverse symbolic calculators did not find anything really close except for $k=1$ for which $$\left(\frac{1}{2};\frac{1}{2}\right)_{\infty }\approx 2 \log (\log (4+e))-1\approx 0.288 788 0950 418 755$$

For large values of $k$, it seems that they can be approximated as $$\left(2^{-k};\frac{1}{2}\right){}_{\infty }=1-2^{(1-k)}+\frac{2^{2(1- k)}}{3} -\frac{2^{3(1- k)}}{21} +\frac{2^{4(1- k)}}{315} -\frac{2^{5(1- k)}}{9765}+\cdots$$ the absolute value of the reciprocals of coefficients corresponding to sequence $A005329$ in $OEIS$ (these numbers seem to be called 2-factorial numbers).

According to this page, the value for $k=1$ is apparently given by $$2^{-7/24}\sqrt[3] {\theta'_1(2^{-1/2})}$$ where appears a Jacobi theta function.

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