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I found this interesting fraction sequence: $$1,\frac{1}{2},\frac{\frac{1}{2}}{\frac{3}{4}}, \frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}} ... $$

This sequence surprisingly converges to $\frac{\sqrt{2}}{2}$

Can someone provide a proof or an idea for this ?

Here's my work. I took the 5th term of the sequence and reciprocated everything to get a 1 2 3 4 5 6 pattern in the manner shown by the arrows. It repeats itself at the dotted edge with the dotted edge being common to the two recurring units.

Here's the INCORRECT image attached.

enter image description here

edit : I am very sorry for switching 11 and 12. This was a silly error on my part. This then changes the pattern and it becomes more confusing now.

here's the correct pattern for the first five terms:

I have also written the rules I have identified.

maybe someone can help me spot a pattern just like @C Squared did.

And I seem to like the pattern that @Stinking Bishop has spotted. The binary one. Maybe those lines work. But we want a general (probably recursive) formula.

Note- I concluded that it converges to $\frac{\sqrt{2}}{2}$ by anaysing the first 7 terms.
p.s. - And i did it by hand because i don't know programming :(

Math Monk
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  • The numerator is defined as $p_1 = 1, p_2 = 4, p_3 = 6$ and $p_{3(n+1)+k} = p_{3n+k}+6$ for each $k=0,1,2$. The denominator is defined similarly... – C Squared Dec 29 '22 at 08:10
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    How do you know that it converges to $\sqrt{2}/2$? Also, are you sure that $11/12$ is the right way round? It seems to me that $\frac{2n+1}{2n+2}$ is reciprocated if and only if $n$ has odd parity when written in binary. If I am right, this breaks the pattern that you and @CSquared have spotted. –  Dec 29 '22 at 08:13
  • In fact, $n$ is in numerator if $n-1$ has even parity and in denominator if $n-1$ has odd parity. I don't know how (if at all!) this helps. –  Dec 29 '22 at 08:19
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    Why don't you erase the first figure ? – Jean Marie Dec 29 '22 at 11:18

1 Answers1

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The answer can be found there, with reference to other nice properties of the Prouhet Thue Morse sequence: https://cs.uwaterloo.ca/~shallit/Papers/ubiq15.pdf

Let $(\epsilon_n)$ be the sequence defined by $\epsilon_0=0$, $\epsilon_{2n}=\epsilon_n$ for every $n \ge 1$, $\epsilon_{2n+1}=-\epsilon_n$ for every $n \ge 0$. The fractions mentionned by Maou are partial products of the infinite product $$P := \prod_{n=0}^{+\infty} \Big(\frac{2n+1}{2n+2}\Big)^{\epsilon_n}.$$ Let $$Q := \prod_{n=1}^{+\infty} \Big(\frac{2n}{2n+1}\Big)^{\epsilon_n}.$$ Grouping terms two by two, one checks that both infinite products are convergent (and positive). One checks that $$PQ = \frac{1}{2}\prod_{n=1}^{+\infty} \Big(\frac{n}{n+1}\Big)^{\epsilon_n} = \frac{Q}{2P},$$ thanks to the relations $\epsilon_{2n}=\epsilon_n$ and $\epsilon_{2n+1}=-\epsilon_n$. The value of $P$ follows. Nothing is known about $Q$.

An amazing result states that $\epsilon_n$ equals $1$ or $-1$ according that the partial product $\prod_{k=0}^{n-1} \Big(\frac{2k+1}{2k+2}\Big)^{\epsilon_k}$ is larger or smaller than $1/\sqrt{2}$. See references given in the link above.

Christophe Leuridan
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    From the help center: "Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the external resource is unreachable or goes permanently offline. Links to other websites should always be helpful, but avoid making it necessary to click on them as much as possible." – Thomas Markov Dec 29 '22 at 18:01
  • @ThomasMarkov I completed my answer. – Christophe Leuridan Dec 29 '22 at 18:56