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Define the well known Riemann-zeta function

$$\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$$

for natural number $n\gt1$

and let $a_1(n)$ be a sequence $a_1(1),a_1(2),a_1(3),\dots$ of the first partial quotients of the simple continued fraction for $\zeta(2n+1)$,

$$\zeta(2n+1)=\{a_0(n);a_1(n),a_2(n),a_3(n),a_4(n),a_5(n),\dots\}$$

The sequence $a_0(n)=1$, and the other sequence $a_1(n)$ increases with n,while the rest of the sequence of partial quotients $a_2(n),a_3(n),a_4(n),\dots$ seem to be random

How do we prove that $\lim_{n\rightarrow\infty} \frac{a_1(n)}{a_1(n+1)}=\frac{1}{4}$

Example

$\zeta(151)=\{1;2854495385411919762116571931558091627803291516,1,4,1,1,\dots\},$ see wolframalpha

and

$\zeta(153)=\{1;11417981541647679048466287742545474611129169915,7,1,15,1,\dots\},$ see wolframalpha

then

$$\frac{a_1(75)}{a_1(76)}=\frac{2854495385411919762116571931558091627803291516}{11417981541647679048466287742545474611129169915}=0.2499999999999999999999999996428\dots$$

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    $\begingroup$ Isn't the 0th partial quotient just '1' for all sufficiently large $n$? Presumably you want the next one. But even there, $\zeta(n) = 1+\frac{1}{2^n}(1+O(c^{-n}))$ for a constant $c$ so the ratio you're looking for can't be right. $\endgroup$ – Steven Stadnicki Sep 13 '17 at 18:42
  • $\begingroup$ @stevenStadnicki: ... and thus $\zeta(2n+1) = 1 + \frac{1}{2^{2n+1}}(1 + O(c^{-n}))$. $\endgroup$ – Hurkyl Sep 14 '17 at 12:17
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This is a straightforward consequence of the asymptotic gap in size between the $\frac1{2^n}$ term in $\zeta(n)$ and the subsequent terms. Note that as $n\to\infty$ we have $\zeta(n)=1+\frac1{2^n}+o(c^n)$ for any $c\gt \frac13$; for concreteness' sake I'll use $c=\frac25$. Because it's a bit easier (and implies the result given) I'll look at the sequence $b_i(n)$ of continued fraction coefficients for $\zeta(n)$ (rather than $\zeta(2n+1)$). Now, $b_0(n)$ is obviously $1$ for all sufficiently large $n$; as noted in comments, we then have $$\begin{array} \ b_1(n) &= \lfloor(\zeta(n)-1)^{-1})\rfloor \\ &=\left\lfloor\left(\left(\frac12\right)^n+o(c^n)\right)^{-1}\right\rfloor \\ &=\left\lfloor\left(\left(\frac12\right)^n\left(1+o((2c)^n)\right)\right)^{-1}\right\rfloor\\ &=\left\lfloor2^n(1+o(d^n))^{-1}\right\rfloor &(\text{writing}\ d=2c)\\ &=\left\lfloor2^n(1-o(d^n))\right\rfloor &(\text{since}\ d=\frac45\lt1)\\ &=2^n(1-o(d^n))\\ \end{array}$$ (Since the $o()$ term here is still more than large enough to 'eat' the maximum difference of $1$ implied by the floor operation). Now, $b_1(n)^{-1} = 2^{-n}(1+o(d^n))$ and $b_1(n+1) = 2^{n+1}(1-o(d^{n+1}))$, so $\frac{b_1(n+1)}{b_1(n)}=2(1+o(d^n))(1-o(d^{n+1}))$ $=2(1+o(d^n))$, and this clearly converges to $2$ as $n\to\infty$.

(To see that this implies the result for $a_1(n)$, note that $a_1(n) = b_1(2n+1)$, so $\frac{a_1(n+1)}{a_1(n)} =\frac{b_1(2n+3)}{b_1(2n+1)}$ $=\frac{b_1(2n+3)}{b_1(2n+2)}\frac{b_1(2n+2)}{b_1(2n+1)}$, so $\lim_{n\to\infty}\frac{a_1(n+1)}{a_1(n)} =\left(\lim_{n\to\infty}\frac{b_1(2n+3)}{b_1(2n+2)}\right)\left(\lim_{n\to\infty}\frac{b_1(2n+2)}{b_1(2n+1)}\right)$ $=2\cdot 2=4$.)

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