An asymptotic series for $\small\prod_{k=n}^\infty\operatorname{sinc}\left({2^{-k}}\pi\right),\,n\to\infty$

Question

Using empirical methods, I conjectured that$^{[1]}$$\!^{[2]}$ $$\small\prod_{k=n}^\infty\operatorname{sinc}\left({2^{-k}}\pi\right)=1-\frac{2\pi^2}9\,4^{-n}+\frac{38 \,\pi ^4}{2025}\,4^{-2n}-\frac{2332\,\pi ^6}{2679 075}\,4^{-3 n}+\frac{265618\,\pi^8 }{10247461875}\,4^{-4 n}+O\!\left(4^{-5 n}\right)\!.$$ Can we prove this? Can we find more coefficients in this expansion, or a general formula for those coefficients?

Answer 1

Writing $\operatorname{sinc} x$ as an infinite product, taking the logarithm and changing the summation order in the triple sum gives $$\ln \prod_{k \geq n} \operatorname{sinc}(2^{-k} \pi) = \sum_{k \geq n} \ln \prod_{l \geq 1} \left( 1 - \frac {2^{-2 k}} {l^2} \right) = -\sum_{k \geq n} \sum_{l \geq 1} \sum_{m \geq 1} \frac 1 m \left( \frac {2^{-2 k}} {l^2} \right)^{\!m} = \\ \sum_{m \geq 1} c_m \frac {2^{-2 m n}} {m!}, \quad c_m = -\frac {2^{2 m} \Gamma(m) \zeta(2 m)} {2^{2 m} - 1}.$$ Then, by Faa di Bruno's formula, $$\prod_{k \geq n} \operatorname{sinc}(2^{-k} \pi) = 1 + \sum_{m \geq 1} \sum_{1 \leq l \leq m} B_{m, l}(c_1, \ldots, c_{m - l + 1}) \frac {2^{-2 m n}} {m!} = \\ 1 + \sum_{m \geq 1} B_m(c_1, c_2, \ldots, c_m) \frac {2^{-2 m n}} {m!},$$ where $B_{m, l}$ and $B_m$ are the Bell polynomials.