How to prove that $\lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{\Pi_{i=1}^n (2^i-1)} < \infty$?

Question

The problem is as follows:

$$\lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{\Pi_{i=1}^n (2^i-1)}$$

I'm stuck on where to go with this. When I plug the problem into Wolfram Alpha, I get this, which I can't decipher, as I'm not familiar with the q-Pochhammer Symbol or its properties.

The only thing I've been able to try is factoring out a $2^n$ from the top and bottom, but this still leaves me stuck on where to go. Help would be appreciated. Thanks in advance.

Answer 1

If you just want to prove it converges (numerically, to some number $\ell \approx 3.46274662$), you can do as follows: since $$ \frac{2^\frac{n^2+n}{2}}{\prod_{i=1}^n (2^i-1)} = \frac{1}{\prod_{i=1}^n \left(1-\frac{1}{2^i}\right)} $$ is is sufficient to show that $a_n := \prod_{i=1}^n \left(1-\frac{1}{2^i}\right)$ converges to some positive number; equivalently, that $b_n := \log a_n$ converges to some real number. Now, $$ \log a_n = \sum_{i=1}^n \log \left(1-\frac{1}{2^i}\right) $$ and this converges by comparison with $-\sum_{i=1}^n \frac{1}{2^i}$.

Answer 2

We have \begin{align} \lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{\prod_{i=1}^n (2^i-1)} &= \lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{\prod_{i=1}^n (2^i(1-2^{-i}))}\\ &= \lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{(\prod_{i=1}^n 2^i)(\prod_{i=1}^n(1-2^{-i}))}\\ &= \lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{2^{\sum\limits_{i=1}^n i}\prod_{i=1}^n(1-2^{-i})}\\ &= \lim_{n \to \infty} \frac{2^\frac{n^2+n}{2}}{2^{\frac{n^2+n}{2}}\prod_{i=1}^n(1-2^{-i})}\\ &= \lim_{n \to \infty} \frac{1}{\prod_{i=1}^n(1-2^{-i})}\\ &= \frac{1}{\prod_{i=1}^\infty(1-2^{-i})} = \frac{1}{\phi(\tfrac{1}{2})}\\ \end{align}

where $\phi$ is the Euler q-series function and since there's no closed form is known for this function, your question can't be further simplified.