How to compute $\prod_{n=1}^{\infty} (1-p^{-n})$ We know it converges for any prime $p$. I just want to know how to compute its exact value:
$$\prod_{n=1}^{\infty} (1-p^{-n})$$
 A: Wolfram MathWorld gives the expression of this product in terms of the q-Pochhammer symbol and the Jacobi theta function. See formulas (46) and (47) in
https://mathworld.wolfram.com/InfiniteProduct.html
A: Let's fix the issue of giving bounds on the infinite product. The pentagonal number theorem gives, after grouping pairs of consecutive terms with the same sign, an alternating series with terms that are decreasing in modulus. So for instance for $c:=\prod _ {n\ge1} (1-2^{-n})$ one has
$$1-\frac{1}{2}-\frac{1}{4}+\frac{1}{32}+\frac{1}{128} -\frac{1}{4096}-\frac{1}{ 32768 } <   c  <  1-\frac{1}{2}-\frac{1}{4}+\frac{1}{32}+\frac{1}{128}$$ that is  $ 0.288787842 < c <  0.2890625, $ in any case larger than $1/4$.
A: The Dedekind eta function is
$$
\eta = q^{1/24}\prod_{k=1}^\infty(1-q^k) = 
e^{\pi i \tau/12}\prod_{k=1}^\infty(1-e^{2\pi i k \tau})
$$
which converges for complex $q$ with $|q| < 1$.  It is often written in terms of a complex argument $\tau$ with $\text{Im}\;\tau > 0$, where $q=e^{2\pi i \tau}$.
The factor $q^{1/24}$ in front gives this desirable number-theoretic properties, but clearly it can be evaluated with that factor if and only if it can be evaluated without.
Some exact values are known for the eta function (see the Wikipedia page).  But not $q=1/p$, $p$ prime.
