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$a_1=\frac{1}{2},a_2=\frac{\frac{1}{2}}{\frac{3}{4}},a_3=\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}$... Prove that $\lim\limits_{n\rightarrow \infty}{a_n}=\frac{\sqrt{2}}{2}$

Notice that if $\frac{i}{i+1}$ is a numerator, then $\frac{i+2^n}{i+1+2^n}$ is a denominator. Each pair of number which is numerator is different from the pair of number that minus $2^n$.

For example, we notice that $a_3=\frac{1}{2}\frac{4}{3}\frac{6}{5}\frac{7}{8}$ and $\frac{1}{2}\rightarrow 00,\frac{3}{4}\rightarrow 01,\frac{5}{6}\rightarrow 10,\frac{7}{8}\rightarrow 11$

I check the result by MATLAB. enter image description here

user
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pchappy
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    What have you attempted thus far? – Mr Pie Apr 16 '18 at 06:02
  • Write $a_n=\prod_{k=1}^{2^n}k^{b_{nk}}$, each exponent $\pm 1$. – J.G. Apr 16 '18 at 06:23
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    Note: For $t_n = 1$ if $n$ occurs in the numerator, and $t_n = 0$ if $n$ occurs in the denominator, then $t_n$ is the Thue-Morse sequence, and it can be constructed starting from 1, and glue the complement of the previous length 2^k segment. That is, 1, 10, 1001, 10010110, .... The limit of the this construction is $(t_n)$. – bonsoon Apr 16 '18 at 06:27
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    @Community Possible duplicate of Tall fraction puzzle. Probably the first and best answered appearance of the problem on SE. – user Apr 16 '18 at 10:06

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