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What is a good/tight upper bound for the infinite product $$\prod_{i=0}^\infty \left(1+\frac{p}{2^i}\right)$$ as $p\to \infty$? Indeed, it's bounded above by $e^{2p}$ so at least its finite, but that is exponential in $p$, and not quite useful. Ideally, one would like a bound in polynomial of $p$ or may quasi-polynomial in the sense of $\exp( (\log p)^n)$ for some $n\ge 1$.

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3 Answers 3

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If we let $$ f(x) = \prod\limits_{n = 0}^\infty {\left( {1 + \frac{x}{{2^n }}} \right)} , $$ then it is easy to see that $f(2x) = (1 + 2x)f(x)$, i.e., $$ \log f(2x) = \log (1 + 2x) + \log f(x). $$ Now for $0<x\le 1$, say, $$ \left| {\log f(x)} \right| = \sum\limits_{n = 0}^\infty {\log \left( {1 + \frac{x}{{2^n }}} \right)} \le \sum\limits_{n = 0}^\infty {\frac{x}{{2^n }}} = 2x. $$ For $x>1$, we find \begin{align*} \left| {\log f(x)} \right| & = \sum\limits_{n = 0}^{\left\lfloor {\log _2 x} \right\rfloor } {\log \left( {1 + \frac{x}{{2^n }}} \right)} + \sum\limits_{n = \left\lfloor {\log _2 x} \right\rfloor + 1}^\infty {\log \left( {1 + \frac{x}{{2^n }}} \right)} \\ & \le (\left\lfloor {\log _2 x} \right\rfloor + 1)\log (1 + x) + \sum\limits_{n = \left\lfloor {\log _2 x} \right\rfloor + 1}^\infty {\frac{x}{{2^n }}} \\ & \le (\left\lfloor {\log _2 x} \right\rfloor + 1)\log (1 + x) + 2 = \mathcal{O}(\log ^2 x). \end{align*} Thus, the Mellin transform $F$ of $\log f$ exists for $-1<\text{Re}(s)<0$ and it satisfies the identity $$ \frac{1}{{2^s }}F(s) = \frac{1}{{2^s }}\frac{\pi }{{s\sin (\pi s)}} + F(s). $$ Then $$ F(s) = \frac{1}{{1 - 2^s }}\frac{\pi }{{s\sin (\pi s)}}, $$ and hence, by Mellin inversion, $$ \log f(x) = \frac{1}{{2\pi {\rm i}}}\int_{ - 1/2 - {\rm i}\infty }^{ - 1/2 + {\rm i}\infty } {\frac{{x^{ - s} }}{{1 - 2^s }}\frac{\pi }{{s\sin (\pi s)}}{\rm d}s} . $$ Assume that $x\ge 1$. If we push the contour through the poles at the non-negative integers and at the points $\frac{{2\pi {\rm i}m}}{{\log 2}}$ (with $m=\pm 1,\pm 2,\pm 3,\ldots$), we deduce $$ \log f(x) = \frac{{\log ^2 x}}{{2\log 2}} + \frac{{\log x}}{2} + \frac{{\log 2}}{{12}} + \frac{{\pi ^2 }}{{6\log 2}} - \frac{1}{x} + \frac{1}{{6x^2 }} - \frac{1}{{21x^3 }} + \ldots\,. $$ (Note that the residue contributions from the imaginary poles cancel each other.) Taking exponentials, we finally have $$\boxed{ f(x) = \exp \!\bigg( \frac{{\log 2}}{{12}} + \frac{{\pi ^2 }}{{6\log 2}} \bigg)\sqrt x \exp\! \bigg( {\frac{{\log ^2 x}}{{2\log 2}}} \bigg)\left( {1 - \frac{1}{x} + \frac{2}{{3x^2 }} - \frac{8}{{21x^3 }} + \ldots } \right)} $$ for $x\ge 1$.

Remark. If one pushes the contour through the pole at $n\ge 1$, the absolute value of the remainder term is at most \begin{align*} \frac{1}{{\left| x \right|^{n + \frac{1}{2}} }}\int_{n + 1/2 - {\rm i}\infty }^{n + 1/2 + {\rm i}\infty } {\frac{{\left| {{\rm d}s} \right|}}{{\left| {s\sin (\pi s)} \right|}}} & = \frac{1}{{\left| x \right|^{n + \frac{1}{2}} }}\int_{n + 1/2 - {\rm i}\infty }^{n + 1/2 + {\rm i}\infty } {\frac{{\left| {{\rm d}s} \right|}}{{\left| s \right|\cosh (\pi {\mathop{\rm Im}\nolimits} (s))}}} \\ & \le \frac{1}{{\left| x \right|^{n + \frac{1}{2}} }}\int_{n + 1/2 - {\rm i}\infty }^{n + 1/2 + {\rm i}\infty } {\frac{{\left| {{\rm d}s} \right|}}{{{\mathop{\rm Re}\nolimits} (s)\cosh (\pi {\mathop{\rm Im}\nolimits} (s))}}} \\ & = \frac{2}{{\left( {n + \frac{1}{2}} \right)\left| x \right|^{n + \frac{1}{2}} }}\int_0^{ + \infty } {\frac{{{\rm d}t}}{{\cosh (\pi t)}}} = \frac{1}{{\left( {n + \frac{1}{2}} \right)\left| x \right|^{n + \frac{1}{2}} }}. \end{align*} Thus, pushing the contour to infinity indeed produces a convergent expansion for $x\ge 1$.

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We can show that $P(x)=\prod_{j=0}^\infty (1+x/2^j)$ satisfies $\log P(x) \approx \log^2 x$ so $e^{A\log^2 x} \le P(x) \le e^{B\log^2 x}$ for some $A,B>0$

$\log P(x)=\sum_{j=1}^{[2\log x]+1}\log (1+x/2^j)+\sum_{j \ge [2\log x]+2}\log (1+x/2^j)=S_1+S_2$

Since $[2\log x]+2>2\log x+1 >\log_2 x+1$ we have that for $j$ in the second sum we have $2^j >2x$ so using that $\log (1+y)=O(y), 0<y<1/2$ we get that $\log (1+x/2^j)=O(x/2^j)=O(1/2^{j-j_0})$ where $j_0=[2\log x]+1$ so $$S_2=O(\sum_{j >j_0}1/2^{j-j_0})=O(1)$$

For $S_1$ we use that each term is at most $\log x$ and there are $[2\log x]+1$ of them so we get that $S_1=O(\log^2 x)$ hence $P(x)<< e^{B\log^2 x}$

For the converse, we take the sums up to $j=[\log x]$ so $2^j \le 2^{\log x}=x^{\log 2}$ so $x/2^j \ge x^{1-\log 2}$ hence $\log (1+x/2^j) > (1-\log 2)\log x$ and since those are about $\log x$ in number and the rest are nonnegative we get $\log P >(1-\log 2)\log^2 x$ so $P >> e^{A \log ^2 x}$

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$$f(p) = \prod\limits_{i = 0}^\infty {\left( {1 + \frac{p}{{2^i }}} \right)}\quad \implies \qquad \log(f(p))= \sum\limits_{i = 0}^\infty \log\Bigg({\left( {1 + \frac{p}{{2^i }}} \right)}\Bigg)$$ Since $$\int \log\Bigg({\left( {1 + \frac{p}{{2^i }}} \right)}\Bigg)\,di=\frac{\text{Li}_2\left(- \frac{p}{{2^i }}\right)}{\log (2)}$$ wee can try Euler-MacLaurin summation and obtain

$$\log(f(p))=\frac{1}{2} \log (p+1)-\frac{\text{Li}_2(-p)}{\log (2)}+\sum_{n=1}^\infty a_n\, \left(\frac p {p+1}\right)^n$$ where the $a_n$ are polynomials in $\log(2)$.

Expanded for an infinite value of $p$ $$\large\color{blue}{\log(f(p))=\frac{2 \pi ^2+\log ^2(2)}{12 \log (2)}+\frac{\log ^2(p)}{2 \log (2)}+\frac{\log (p)}{2}-\frac 1 p+\frac 1 {6p^2}-}$$ $$\large\color{blue}{\frac 1 {21p^3}+\frac 1 {60p^4}-\frac 1 {155p^5}+\frac 1 {383p^6}+O\left(\frac{1}{p^7}\right)}$$ which is almost "exact" (it even seems to be an uper bound).

For $p=10^6$, the above gives $\color{red}{147.02105276800}372$.

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  • $\begingroup$ Your expansion agrees with mine and I showed that it is exact for $p\ge 1$. $\endgroup$
    – Gary
    Dec 20, 2023 at 6:22
  • $\begingroup$ @Gary. I know that but you solution is so much more elegant. Cheers :-) $\endgroup$ Dec 20, 2023 at 6:40

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