<s>[INST] Prove $\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$ Question:

Prove or disprove
$$I=\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$$

I know 
\begin{align}\frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=&\left(1+\dfrac{1}{2^{2^n}}\right)\left(1+\dfrac{1}{2^{2^n}+2}\right)\\&\cdots\left(1+\dfrac{1}{2^{
2^n+1}}\right)\end{align}
so
$$\lim_{n\to\infty}\left(1+\dfrac{1}{2^{2^n}}\right)\left(1+\dfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\dfrac{1}{2^{
2^n+1}}\right)=\sqrt{2}\ ?$$
I feel this result is very surprising. This problem comes from Chris's sis.
and I use wolfram,limit wofl can't find it
I often use this theta function and is this true? Thank you.
 [/INST] The logarithm of the expression under the limit can be rewritten as
$$\sum_{k=0}^{2^{2^n-1}}\ln\left(1+\frac{1}{2^{2^n}+2k}\right)=\sum_{k=0}^{2^{2^n-1}}\frac{1}{2^{2^n}+2k}+O\left(2^{-2^n}\right).$$
Denoting $N=2^{2^n-1}$, it is easy to see that the limit of the logarithm can be computed as the limit of a Riemann sum:
$$\frac{1}{N}\sum_{k=0}^N\frac{1}{2\left(1+\frac{k}{N}\right)}\stackrel{N\rightarrow\infty} \longrightarrow \frac12\int_0^1\frac{dx}{1+x}=\ln\sqrt2.$$
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