For $x \in \mathbb{R}^+$, let $\{x\} = x - \lfloor x \rfloor$ denote the fractional part of $x$. Let $k \in \mathbb{N}$. Show that
$2^{\{k \log_2(3)\}} < \dfrac{2}{1 + 2^{-k}}$
for $k > 1$.
This inequality seems very true, as in there should be a much tighter bound. It has something to do with the properties of $\log_2(3)$, probably the fact that $2$ and $3$ are coprime integers. It is closely related to the ergodic circle rotation with $\alpha = \log_2(3) \pmod{1}$.
Any help would be greatly appreciated.
Edit: It seems doable to use the equality
$$k \log_2(3/2) = \dfrac{2k}{\log(2)} \sum_{n=1}^\infty \frac{1}{(2n-1) \cdot 5^{2n-1}}$$
For any $k$, there is an $m$ such that $k < (2m-1) \cdot 5^{2m-1}$. Then,
$$\{k \log_2(3/2)\} = \left\{ \dfrac{2}{\log(2)} \left( \sum_{n=1}^{m-1} \frac{k \pmod{(2n-1) \cdot 5^{2n-1}}}{(2n-1) \cdot 5^{2n-1}} + \sum_{n=m}^\infty \frac{k}{(2n-1) \cdot 5^{2n-1}} \right) \right\}$$
The sum on the right shrinks very fast. We should be able to pick an $m$ so that this contribution is negligible, less than some $\varepsilon$. The sum on the left shouldn't get very large because it gets "all mixed up." There is some value of $2 \leq k \leq (2m-3)\cdot 5^{2m-3}$ that maximizes it with a sum no greater than $1 - [(2m-3)\cdot 5^{2m-3}]^{-1}$.
So, we just need to show that the sum on the right shrinks fast enough so that we can pick an $m$ small enough so that
$$\log_2(1 + 2^{-k}) < [(2m-3)\cdot 5^{2m-3}]^{-1} - \varepsilon$$
Then, we would have
$$\{k \log_2(3)\} < 1 - \log_2(1 + 2^{-k})$$
as desired. This seems very possible because $m$ should grow exponentially slower than $k$. At least, that's the idea I've got so far.
Edit: I forgot about the $1/\log 2$ term, which completely throws off this entire argument :/