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.
Best Answer
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.