How to find the constant $C$ such that $f(x)\geq Cx$

Question

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 .

Answer 1

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 $Cx$.

Answer 2

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}

Answer 3

If, as given by @metamorphy, we use $$C= \prod_{k=1}^\infty \left(1- \frac{1}{2^k (k+1)} \right)$$ Taking logarithms as @Ragno did $$\log(C)=-\sum_{m=1}^\infty \frac{ \Phi \left(2^{-m},m,2\right)}{m\, 2^{m}}=-\sum_{m=1}^\infty \frac {2^m\, \text{Li}_m\left(2^{-m}\right)-1}m $$ which converges very fast since $$\frac{2^m\, \text{Li}_m\left(2^{-m}\right)-1}{m}\sim \frac{2^{-2 m}}{m}$$

All of this leads to $$\log(C) = -0.428248658776493709190419287\cdots$$ for which the $ISC$ suggests $$\color{blue}{\large-\left(\frac{\sqrt{5}}{8 \,\phi }\right)^{\frac{\sqrt{2+\sqrt{3}}}{4}}}$$ which is in an absolute error of $3.23 \times 10^{-8}$.

$$C=e^{\log(C)}=0.651649356319290075542872424\cdots$$ Amazing is that the next number given in the $ISC$ is $$\color{blue}{\large\frac{2}{\phi }\,{ \Omega ^{\Gamma \left(\frac{5}{6}\right)}}}$$ which is in an absolute error of $1.56 \times 10^{-8}$.

Edit

Hoping to please the OP in this holiday season, $$\color{green}{C\sim\frac{2473 }{2835}\,\,\binom{\gamma }{\gamma !}+\frac{703 }{2835}\,\,\binom{\gamma !}{\gamma }+\frac{479}{945}\,\, \binom{\gamma !}{\log (\gamma )}-}$$ $$\color{green}{\frac{143}{810} \,\,\binom{\log (\gamma )}{\gamma !}-\frac{15 }{14}\,\,\binom{\gamma }{\log (\gamma )}+\frac{2129 }{5670}\,\,\binom{\log (\gamma )}{\gamma }}$$ gives all the figures.

Answer 4

Just to see some interesting pattern :

Always with a computer and if I don't make mistake we have

$$\left|\exp\left(1-\prod_{k=1}^{1000}\left(1-\frac{1}{2^{k}(k+1)}\right)\right)-\sqrt{2}-1/2^{9}-1/2^{11}-1/2^{14}-1/2^{16}+1/2^{18}-1/2^{20}-1/2^{21}-1/2^{24}+1/2^{27}+1/2^{30}-1/2^{33}+1/2^{36}+1/2^{38}+1/2^{39}-1/2^{41}-1/2^{42}-1/2^{43}-1/2^{47}-1/2^{49}+1/2^{51}+1/2^{52}-1/2^{55}+1/2^{57}-1/2^{60}-1/2^{63}-1/2^{64}+1/2^{66} \right|<1*10^{-21}$$

plugging it in oeis I haven't found any things .

We have again with a computer :

$$|\exp\left(3\left(\pi-e\right)^{2}-\frac{1}{\pi-e}\prod_{k=1}^{1000}\left(1-\frac{1}{2^{k}(k+1)}\right)\right)+\frac{1}{\pi^{2\pi}}-e^{-1}-\frac{1}{\pi^{e\pi}}-\frac{1}{\pi^{3\pi}}|<4*10^{-7}$$

I can add some other but I let the most saying .