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I stumbled upon something rather interesting today while noodling around. Given that this general formula for all $\zeta(x)$ produces a continued fraction, does it therefore stand that all $\zeta(x)$ are irrational?
$\zeta(x)$= $1+ \cfrac{1^{2x}}{ 2^x-\cfrac{2^{2x}}{K_2 + K_3-\cfrac{3^{2x}}{K_3 + K_4-\cfrac{4^{2x}}{K_4 + K_5-\ddots}}}}$

Where $K_n$ is the denominator of the $n^{th}$ Riemann (e.g. $\zeta(3)$, $K_n$:{1,8,27,64 $\dots$}, otherwise noted as $n^x$). $$-$$ Notice, that $K_n$ + $K_{n+1}$ in Apery's $\zeta(3)$ is nothing more than https://oeis.org/A005898 from 35 onwards, or https://oeis.org/A001844 for $\zeta(2)$

P_Corfield
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  • No, this is not enough to tell you $\zeta(x)$ is irrational. Continued fractions will tell you the result is irrational in case all numerators are $1$ and all denominators are positive integers. – GEdgar Oct 22 '19 at 23:40
  • Where does your continued fraction come from, the only one I know is thisone, for asymptotic reasons I doubt there are other expansions of $\zeta(s)$ in such continued fraction of $a^{-s}$ terms – reuns Oct 22 '19 at 23:42
  • @reuns - I just made it up today. I tried it for $\zeta(x)$ up to 16 and it worked like a charm – P_Corfield Oct 22 '19 at 23:48
  • Then show us your proof, how did you come to it, and define $K_n$ correctly and for which $x$ it is supposed to hold – reuns Oct 22 '19 at 23:52
  • @reuns - It appears to hold for all x. The process is incredibly easy. Start with $\zeta(3)$ for example. 1 + $\frac{1}{8+ r}$ = 1 + $\frac{1}{8}$ + $\frac{1}{27}$, Solve for r. Repeat, onward until the pattern is uncovered. – P_Corfield Oct 23 '19 at 00:15
  • $K_n = n^x$. Why not write that? Notice the signs were edited recently. Numerical checks show that this works. Even $\zeta(3/2)$ came out right. – GEdgar Oct 23 '19 at 00:16
  • @GEdgar, sure - was just trying to make it easiest to understand – P_Corfield Oct 23 '19 at 00:18
  • @reuns. Probably not, but r in this example would solve to $\frac{-2^{6}}{35}$, the next term is $\frac{-3^{6}}{91}$ – P_Corfield Oct 23 '19 at 00:28
  • For $s > 1$, $\zeta(s)=\sum_{n=1}^\infty n^{-s}=\sum_{n=1}^\infty \prod_{j=2}^n (\frac{j-1}{j})^s=\sum_{n=1}^\infty \prod_{j=2}^n (1-j^{-1})^s$, with $\zeta_N(s)=\sum_{n=N}^\infty \prod_{j=N+1}^{n-1} (1-j^{-1})^s$ then $\zeta_{N-1}(s)=1+(1-N^{-1})^s\zeta_N(s)=\frac1{1-\frac{(1-N^{-1})^s}{(1-N^{-1})^s+\frac1{\zeta_N(s)}}}$ which gives a continued fraction – reuns Oct 23 '19 at 01:14
  • @P_Corfield Here is another continued fraction representation of $\zeta(s)$. – Neves Oct 23 '19 at 08:51
  • Here we have an instance of Euler's continued fraction formula https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula , for example $${\frac {1} {u_{1} } } +{\frac {1} {u_{2} } } +{\frac {1} {u_{3} } } +\cdots +{\frac {1} {u_{n} } } ={\frac {1} {u_{1} -} } {\frac {u_{1} ^{2} } {u_{1} +u_{2} -} } {\frac {u_{2} ^{2} } {u_{2} +u_{3} -} } \cdots {\frac {u_{n-1} ^{2} } {u_{n-1} +u_{n} } } ,$$ – GEdgar Oct 23 '19 at 09:54
  • @GEdgar - thank you for this. – P_Corfield Oct 23 '19 at 11:11
  • @GEdgar - Although it's not exactly spelled out the same, I agree it is of that form. Interesting, the GCD ($u_{n-1}$ + $u_{n}$, $u_{n-1} ^{2}$) is 1, so that is why I was under the impression this might play a role in determining irrationality – P_Corfield Oct 23 '19 at 12:20

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