Problem:
The following expression is given: $$f(x) = \prod_{n=1}^\infty \sqrt[\displaystyle{2^n}]{x+2^{n-1}} = \prod_{n=1}^\infty \left(x+2^{n-1}\right)^{\displaystyle0.5^n}$$
How fast does it grow in $x$? To be clear, I would like to find good lower and upper bounds $L(x)$ and $U(x)$, such that $$L(x) << f(x) << U(x)$$ Optimally, an exact growth rate would be the best: $f(x) \sim \Theta(F(x))$ for some $F:\Bbb{R}\to\Bbb{R}$.
Used notations:
- If $g(x)$ is a real function, then $\Theta(g(x))$ is a family of functions such that $\forall f(n) \in \Theta(g(n))\;\exists k_{1}>0\;\exists k_{2}>0\;\exists n_{0}\;\forall n>n_{0}\;k_{1}\cdot g(n)\leq f(n)\leq k_{2}\cdot g(n)$
- For real functions $f(x)$ and $g(x)$: $f << g$ or $f(x) << g(x)$, if $\lim_{x\to\infty}\frac{f(x)}{g(x)} = 0$
- For real functions $f(x)$ and $g(x)$: $f \sim g$ or $f(x) \sim g(x)$, if $\lim_{x\to\infty}\frac{f(x)}{g(x)} = 1$
Assumption:
I calculated some values for $f(x)$:
\begin{array} {|r|r|} \hline x & f(x) \\ \hline 0 & 2 \\ \hline 1 & 2.5 \\ \hline 10 & 3 \\ \hline 100 & 4.4 \\ \hline 1000 & 5.9 \\ \hline 10^5 & 9.3 \\ \hline 10^7 & 11.77 \\ \hline \end{array}
From this it would seem that $f(x) \sim \log_a(x+b) + c$ for some $a,b,c \in \Bbb{R}$. However, I have not been able to find values for $a,b,c$. Any help would be appreciated!
Try taking logarithms.
$$\begin{align} L&:=\log\prod_{n=1}^\infty \sqrt[{2^n}]{x+2^{n-1}}\\ &=\sum_{n=1}^\infty2^{-n}\log\left(x+2^{n-1}\right)\\ &=\sum_{n=1}^\infty2^{-n}\log\left(2^{n-1}\left(1+\frac{x}{2^{n-1}}\right)\right)\\ &=\sum_{n=1}^\infty2^{-n}(n-1)\log2+ \sum_{n=1}^\infty2^{-n}\log\left(1+\frac x{2^{n-1}}\right) \end{align}$$
The first sum can be computed explicitly, and it should be easy to bound the second using the power series for $\log(1+x)$. Then you have a bound on $e^L$.