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 .

The claim is that $f(x)\ge Cx$ for any $x>0$, where
$$
C= \prod_{k=1}^\infty\left( {1 - \frac{1}{{2^k (k + 1)}}} \right) =0.651649356319290\ldots,
$$
as given by @metamorphy in the comments. Indeed
\begin{align*}
\log f(x) & = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {1 + \sum\limits_{n = 1}^k {\frac{1}{{k \cdot 2^n }}} } \right)}  = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right)} 
\\ &  = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {\frac{{k + 1}}{k}} \right)}  + \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {\frac{{k + 1}}{k}} \right)} \right]} 
\\ &  = \log (\left\lfloor x \right\rfloor  + 1) + \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right)} \right]} 
\\ &  \ge \log x + \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right)} \right]} 
\\ &  \ge \log x + \sum\limits_{k = 1}^\infty  {\left[ {\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right)} \right]} 
\\ & =\log x+ \sum\limits_{k = 1}^\infty  {\log \left( {1 - \frac{1}{{2^k (k + 1)}}} \right)} 
\\ & = \log x + \log C,
\end{align*}
since
$$
\log \left( {1 + \frac{{1 - 2^{ - k} }}{k}} \right) - \log \left( {1 + \frac{1}{k}} \right) \le 0
$$
for any $k\ge 1$. It is readily seen that
$$
\lim_{x\to +\infty}\frac{f(x)}{x}=C,
$$
i.e., $C$ is the best possible such constant.
Note that $C(\left\lfloor x \right\rfloor+1)$ is a better approximation (and lower bound) to $f(x)$ than $Cx$.
