How to find the constant $C$ such that $f(x)\geq Cx$ Problem :
Define for strictly positive $x$ :
$$f\left(x\right)=\left(\prod_{k=1}^{\operatorname{floor}\left(x\right)}\left(1+\sum_{n=1}^{k}\frac{1}{k\cdot2^{n}}\right)\right)$$
Does there exists a constant $C$ such that :
$$f(x)\geq Cx$$
I think definitely yes and $C\simeq 0.6516...$
Currently Symbolic calculator don't find  it .
As side notes we have :
$$-\ln\left(2\right)+\sum_{k=1}^{\infty}\frac{1}{k\cdot2^{k}}=0$$
$$\frac{1}{2}\left(1-3\ln\left(\frac{4}{3}\right)\right)+\left(+\sum_{k=1}^{\infty}\frac{1}{k\left(k+1\right)\cdot2^{1+2k}}\right)=0$$
and so on .
Edit 26/12/2022 :
we have with a computer :
$$\left|\exp\left(1-\prod_{k=1}^{1000}\left(1-\frac{1}{2^{k}(k+1)}\right)\right)-\sqrt{2}-\frac{1}{2^{9}}-\frac{1}{2^{11}}-\frac{1}{2^{14}}-\frac{1}{2^{16}}+\frac{1}{2^{18}}-\frac{1}{2^{20}}-\frac{1}{2^{21}}-\frac{1}{2^{24}}+\frac{1}{2^{27}}+\frac{1}{2^{30}}-\frac{1}{2^{33}}+\frac{1}{2^{36}}+\frac{1}{2^{38}}+\frac{1}{2^{39}}-\frac{1}{2^{41}}\right|<4*10^{-13}$$
I don't know if we can go like this at infinity .
Question :
How to show the existence and does $C$ admits a closed form ?
Thanks a lot .

Note that
\begin{equation}
       f(x)=\prod_{k=1}^{\left \lfloor x \right \rfloor} \left(1+ \frac{1}{k}\sum_{j=1}^k \frac{1}{2^j}\right)= \prod_{k=1}^{\left \lfloor x \right \rfloor} \left( 1+ \frac{1-2^{-k}}{k} \right)
\end{equation}
Therefore,
\begin{align}
     \ln f(x) =& \sum_{k=1}^{\left \lfloor x \right \rfloor} \ln \left(1+ \frac{1-2^{-k}}{k} \right) \approx \sum_{k=1}^{\left \lfloor x \right \rfloor} \frac{1-2^{-k}}{k}= H_{\left \lfloor x \right \rfloor}-\sum_{k=1}^{\left \lfloor x \right \rfloor} \frac{1}{k 2^k}\\ &\geq  H_{\left \lfloor x \right \rfloor}- \sum_{k=1}^{\infty} \frac{1}{2^k}= H_{\left \lfloor x \right \rfloor}-1
\end{align}
where $H_n$ denotes the harmonic series, and the approximation follows from a Taylor expansion. Next as $H_n\approx \ln n +\gamma$, where $\gamma\approx 0.5772$ is the Euler–Mascheroni constant,
\begin{equation}
     f(x)\geq \frac{1}{e} e^{H_{\left \lfloor x \right \rfloor}} \approx e^{\gamma-1} \left \lfloor x \right \rfloor \geq  e^{\gamma-1} (x-1) \approx 0.6552 (x-1)
\end{equation}
