TITLE: Expressing the Riemann Zeta function in terms of GCD and LCM
QUESTION [27 upvotes]: 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.

REPLY [2 votes]: 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.