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

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.

Answer 1

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: $$\frac{1}{N}\sum_{k=0}^N\frac{1}{2\left(1+\frac{k}{N}\right)}\stackrel{N\rightarrow\infty} \longrightarrow \frac12\int_0^1\frac{dx}{1+x}=\ln\sqrt2.$$

Answer 2

$$\log \left(1+\tfrac{1}{2^{2^n}}\right)\left(1+\tfrac{1}{2^{2^n}+2}\right)\cdots\left(1+\tfrac{1}{2^{ 2^n+1}}\right) = \log \left(1+\tfrac{1}{2^{2^n}}\right)+ \log \left(1+\tfrac{1}{2^{2^n}+2}\right)+\cdots + \log\left(1+\tfrac{1}{2^{ 2^n+1}}\right)$$

Expanding on Alex' idea let $t = 2^{2^n}$ which is getting very large. Notice we get only even numbers:

$$ \log \left(1+\tfrac{1}{t}\right)+ \log \left(1+\tfrac{1}{t+2}\right)+\cdots + \log\left(1+\tfrac{1}{2t}\right) \approx \frac{1}{t} + \frac{1}{t+2}+\cdots + \frac{1}{2t} $$

We get a Riemann sum, but only half of the terms:

$$ \frac{1}{t}\big[1 + \frac{1}{1+\tfrac{2}{t}}+\cdots + \frac{1}{2}\big] \approx \frac{1}{2} \int_1^2 \frac{dt}{t} = \frac{\log 2}{2} = \log \sqrt{2} $$

Answer 3

Wolfram can actually find it: http://www.wolframalpha.com/input/?i=lim%28n+to+infinity%29+prod%28k+%3D+0+to+%282%5E%282%5En+-+1%29+%2B+1%29%29+%281+%2B+1%2F%282%5E2%5En+%2B+2k%29%29 in all it's glory.

I find that condensing the search formula for Wolfram and not relying on its pattern recognition tends to get you better results. In your search query, it probably thought that the top and bottom was identical when finding a pattern.