Expressing the Riemann Zeta function in terms of GCD and LCM

Question

Is the following claim true: Let $\zeta(s)$ be the Riemann zeta function. I observed that as for large $n$, as $s$ increased,

$$ \frac{1}{n}\sum_{k = 1}^n\sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)}{\text{lcm}(k,i)}\bigg)^s \approx \zeta(s+1) $$

or equivalently

$$ \frac{1}{n}\sum_{k = 1}^n\sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)^2}{ki}\bigg)^s \approx \zeta(s+1) $$

A few values of $s$, LHS and the RHS are given below

$$(3,1.221,1.202)$$ $$(4,1.084,1.0823)$$ $$(5,1.0372,1.0369)$$ $$(6,1.01737,1.01734)$$ $$(7,1.00835,1.00834)$$ $$(9,1.00494,1.00494)$$ $$(19,1.0000009539,1.0000009539)$$

Note: This question was posted in MSE. It but did not have the right answer.

Answer 1

Let me denote your LHS by $f(n,s)$. For fixed even $n$ I shall show that $f(n,s)-1\sim\zeta(s+1)-1$ as $s\to\infty$, that is, $$\lim_{s\to\infty}\frac{f(n,s)-1}{\zeta(s+1)-1}=1.$$ This result nicely expresses your numerical observations, which show that the parts after the decimal point seem to be asymptotically the same.

On one hand, we have $\zeta(s+1)-1=2^{-s-1}+3^{-s-1}+\dots$. The terms after the second can be estimated from above by the integral $\int_2^\infty x^{-s-1}dx=\frac{2^{-s}}{s}$, so we see that $\zeta(s+1)-1\sim 2^{-s-1}$.

On the other hand, among pairs $(k,i)$ with $1\leq k\leq n,1\leq i\leq k$, the expression $\frac{\gcd(k,i)}{\operatorname{lcm}(k,i)}$ is equal to $1$ for exactly $n$ pairs $(k,k)$, and is equal to $2^{-1}$ for exactly $n/2$ pairs $(2k,k)$. All other terms, of which there are certainly fewer than $n^2$, are at most $3^{-1}$. Therefore we find $$f(n,s)=\frac{1}{n}\left(n\cdot 1+\frac{n}{2}\cdot 2^{-s}+O(n^23^{-s})\right)=1+2^{-s-1}+o(2^{-s})$$ proving $f(n,s)-1\sim 2^{-s-1}$. It follows that $f(n,s)-1\sim\zeta(s+1)-1$, as we wanted.

Let me emphasize that in the above calculation it was crucial that $n$ was even. If $n$ is odd, then we instead only get $\frac{n-1}{2}$ pairs $(2k,k)$ and the asymptotics get slightly skewed - we then get $f(n,s)-1\sim\frac{n-1}{n}(\zeta(s+1)-1)$. For large $n$ the difference is however, pretty negligible.

Answer 2

A variety of formulas of this type (in the sense of a relation between $\zeta(s)$ and a sum over gcd or lcm) has been derived by Titus Hilberdink and László Tóth in On the average value of the least common multiple of k positive integers (2016), see also On the distribution of the greatest common divisor by Diaconis and Erdȍs. I quote

$$\sum_{i,k=1}^n \big(\text{lcm}(k,i)\big)^s=\frac{\zeta(s+2)}{\zeta(2)}\frac{n^{2s+2}}{(s+1)^2}+{\cal O}(n^{2s+1}\log n),$$ $$\sum_{i,k=1}^n \big(\gcd(k,i)\big)^s=\left(\frac{2\zeta(s)}{\zeta(s+1)}-1\right)\frac{n^{s+1}}{s+1}+{\cal O}(n^{s}\log n),$$ $$\sum_{i_1,i_2,\ldots i_s=1}^n \gcd(i_1,i_2,\ldots i_s)=\frac{\zeta(s-1)}{\zeta(s)}n^s+{\cal O}(n^{s-1}),\;\;s\geq 4.$$

The earliest reference for such series is Ernest Cesàro, Étude moyenne du plus grand commun diviseur de deux nombres (1885).

Answer 3

Your summand is symmetric with respect to $k$ and $i$:

$$f(n,s) = \frac{1}{n}\sum_{k = 1}^n \sum_{i = 1}^{k} \bigg(\frac{\gcd(k,i)}{\operatorname{lcm}(k,i)}\bigg)^s$$

We can sum along skew diagonals to evaluate the sum. That is, we can convert $(k,i)$ to polar form $(\sqrt{k^2 + i^2}, tan^{-1}\frac{i}{k})$. By symmetry of the summand, the rays from the origin have the same value.

That is, when $\theta = tan^{-1}\frac{i}{k}$, $i = k \tan\theta$. We can vary $\theta$ from $[0, \frac{\pi}{4}]$. The $\gcd(j,j\tan \theta)$ is independent of $j$ but depends on $\theta$, we can therefore use $n$:

$$\frac{1}{n}\int_{0}^{\frac{\pi}{4}} \sum_{j=0}^{n} \left[\frac{\gcd(j,j\tan\theta)}{\operatorname{lcm}(j,j\tan\theta)}\right]^{s} d\theta = \int_{0}^{\frac{\pi}{4}} \left[\frac{\gcd(n,n\tan\theta)}{\operatorname{lcm}(n,n\tan\theta)}\right]^{s} d\theta = \sum_{k=0}^{n} \frac{1}{k^{s+1}} \rightarrow \zeta(s+1)$$

The polar integral(The integral is not continuous on $\theta$ but over the irrational($\tan\theta = \frac{i}{k}$ implies $\theta$ is irrational) values that correspond to the rays) to the sum is calculated easily enough because the values along each ray is constant w.r.t to $j$. Every ray corresponds to one of the values of $\frac{1}{k^s}$ but they have a weight of $\frac{1}{k}$.

E.g., the ray that accumulates $1$ has $\theta = \tan^{-1}(1) = \frac{\pi}{4}$, for $\frac{1}{2^s}$ it is $\theta = \tan^{-1}(2)$ but it has $\frac{1}{2}$ the density of $1$. Similarly for all the others.

If you are having trouble following this, simply look at $\frac{\operatorname{lcm}(k,i)}{\gcd(k,i)}$ in "polar" form, I'll make it easy for you(the text format obscures the patterns but they are there):

\begin{matrix} \color{green}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \\ \color{blue}2 & \color{green}1 & 6 & 2 & 10 & 3 & 14 & 4 & 18 & 5 & \\ \color{red}3 & 6 & \color{green}1 & 12 & 15 & 2 & 21 & 24 & 3 & 30 & \\ 4 & \color{blue}2 & 12 & \color{green}1 & 20 & 6 & 28 & 2 & 36 & 10 & \\ 5 & 10 & 15 & 20 & \color{green}1 & 30 & 35 & 40 & 45 & 2 & \\ 6 & \color{red}3 & \color{blue}2 & 6 & 30 & \color{green}1 & 42 & 12 & 6 & 15 & \\ 7 & 14 & 21 & 28 & 35 & 42 & \color{green}1 & 56 & 63 & 70 & \\ 8 & 4 & 24 & \color{blue}2 & 40 & 12 & 56 & \color{green}1 & 72 & 20 & \\ 9 & 18 & \color{red}3 & 36 & 45 & 6 & 63 & 72 & \color{green}1 & 90 & \\ 10 & 5 & 30 & 10 & \color{blue}2 & 15 & 70 & 20 & 90 & \color{green}1 & \\ \end{matrix}

If you look you can see $k$th ray which has the constant value $\frac{\color{green}1}{k^s}$(displayed as just $k$ in the table) but they repeat at a rate of $\frac{\color{green}1}{k}$ along the ray.

Alternatively, if write the table in polar coordinates(we are rotating coordinate space 45 degree's) we get \begin{matrix} \color{green}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & 0 & \color{red}3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & 0 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & \color{red}3 & 0 & 0 & 6 & 0 & 0 & 0 & 0 & \\ \color{green}1 & 0 & 0 & 0 & 0 & 0 & 7 & 0 & 0 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 4 & 0 & 0 & 0 & 8 & 0 & 0 & \\ \color{green}1 & 0 & \color{red}3 & 0 & 0 & 0 & 0 & 0 & 9 & 0 & \\ \color{green}1 & \color{blue}2 & 0 & 0 & 5 & 0 & 0 & 0 & 0 & \color{green}10 & \\ \end{matrix}

Where now the $k$th column is the "$k$th" ray. I.e., the first column in the above table corresponds to the diagonals/rays in the table above it.