# Finding closed form for a product

I have no idea how to solve the below product whose closed form I need to solve a problem, can anyone at the very least guide me to a solution or give me a source to check?

$$\prod_{k=1}^{n}\dfrac{2^k-1}{2^k}$$

• Have you calculated the result for small cases like $n = 1, 2, 3$ and seen what you get? Maybe there is a simple pattern you can try to generalize. – Arthur May 20 at 13:56
• The result is 0 when n≥0, because for the index k=0 we have $2^0-1$ in the numerator. May you have meant for the index to start at k=1? – dxdydz May 20 at 13:57
• Sorry index starts from 1 I edited it. For small cases, the denominator seems very easy but the numerator is what bothers me. It goes 1 3 21 315, I assume there might not be a closed form at all. – Stefan M. May 20 at 13:59
• Well, the denominator is clearly just $2^{n(n+1)/2}$. So yes, it's the numerator which seems troublesome. – Arthur May 20 at 13:59
• A numerical experiment yields an approximate value of 0.288788095088. This product does converge, but this funky constant brings nothing special to mind. – ncmathsadist May 20 at 14:01

\begin{align*} \prod_{k=1}^n\frac{2^k-1}{2^k} &=\frac{1}{2^{n(n+1)/2}}\prod_{k=1}^n\left(2^k-1 \right ) \\ &=\frac{(-1)^n}{2^{n(n+1)/2}}\prod_{k=1}^n\left(1-2^k \right ) \\ &=\frac{(-1)^n}{2^{n(n+1)/2}}\prod_{k=0}^{n-1}\left(1-2^{k+1} \right ) \\ &=\frac{(-1)^n}{2^{n(n+1)/2}}(2;\,2)_n. \end{align*}
$$\prod_{k=1}^n\frac{2^k-1}{2^k}=\left(\frac{1}{2};\,\frac{1}{2}\right)_n.$$