Closed form for the infinite product $S_n=\Pi_{i=1}^{\infty} (1+\frac{1}{2^i})$

Question

Consider the infinite product $S_n=\Pi_{i=1}^{n} (1+\frac{1}{2^i})$. The original problem was to prove that $\Pi_{i=1}^{n} (1+\frac{1}{2^i}) < \frac{5}{2} \ \forall \ n \in \mathbb{Z^+}$. This can be done by writing $S_{n+1}-S_n$ as $\frac{S_n}{2^{n+1}}$, then conjecturing that $S_n - S_1< 1$ and proving this claim by induction - the key being to write $S_{k+1}-S_1$ as the telescoping sum $(S_{k+1}-S_k)+(S_k - S_{k-1}) + ... + (S_2-S_1)$. But since $S_1 =\frac{3}{2}$, we have $S_n < 1+\frac{3}{2} =\frac{5}{2}$, so we are done.

However, according to software, $S_n$ seems to converge to the value $2.3842310290313...$ as $ n \rightarrow \infty$. I wonder if there is a closed form for this special number, and how do we prove that $S_n$ actually converges to this value? Because the result proven in the problem seems considerably weaker than this...

Answer 1

The exact constant is OEIS A079555. There is no closed form in the usual sense, but the number can be represented using the $q$-Pochhammer symbol $(-1/2;1/2)_\infty$.

The proven bound of $2.5$ is already quite tight, in my opinion.

Answer 2

@Parcly Taxel provided the answer.

We can get some reasonable approximation writing $$S_n=\prod_{i=1}^{n} \left(1+\frac{1}{2^i}\right)=\frac{85817142703524375}{36028797018963968}\prod_{i=11}^{n} \left(1+\frac{1}{2^i}\right)$$ Now, taking the logarithm of the product and using the expansion $$\log(1+a)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k} a^k$$ we end with $$S_n=85817142703524375\times 2^{-\left(2^{-n}+\frac{56319}{1024}\right)}$$ which gives a limit equal to $$S_\infty=\frac{85817142703524375}{18014398509481984\times 2^{1023/1024}}\approx 2.3835$$