A good upper bound for the infinite product $\prod_{i=0}^\infty (1+p/2^i)$

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 $$n\ge 1$$.

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 $$0, say, $$\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 $$x\ge 1$$.

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 $$\log (1+y)=O(y), 0 we get that $$\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 $$P >> e^{A \log ^2 x}$$

$$f(p) = \prod\limits_{i = 0}^\infty {\left( {1 + \frac{p}{{2^i }}} \right)}\quad \implies \qquad \log(f(p))= \sum\limits_{i = 0}^\infty \log\Bigg({\left( {1 + \frac{p}{{2^i }}} \right)}\Bigg)$$ 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 $$\color{red}{147.02105276800}372$$.

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