Relationship between GCD, LCM and the Riemann Zeta function

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

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

Question: Is the LHS asymptotic to $\zeta(s+1)$ ?

Update: I have posted this in MO since it was open in MSE.

Answer

With $(A,B) = (ga,gb), \gcd(a,b)=1$ then

$$\sum_{A,B, \gcd(A,B) \le G} \frac{\gcd(A,B)^s}{\mathop{\rm lcm}(A,B)^s} = \sum_{g\le G} \sum_{a,b, \gcd(a,b)=1}\frac{\gcd(ag,bg)^s}{\mathop{\rm lcm}(ag,bg)^s}$$

$$= \sum_{g\le G} \sum_{a,b, \gcd(a,b)=1}\frac{g^s}{(abg)^s} = G \sum_{a,b, \gcd(a,b)=1}\frac{1}{(ab)^s}$$
$$ = G \sum_d \mu(d)\sum_{u,v}\frac{1}{(d^2uv)^s}= G (\sum_d \mu(d)d^{-2s})(\sum_uu^{-s})(\sum_v v^{-s}) = G \frac{\zeta(s)^2}{\zeta(2s)}$$

As $s \to \infty$ it $\to G \zeta(s)^2 \approx G\zeta(s+1)$

Attribution
Source : Link , Question Author : Nilotpal Sinha , Answer Author : Greg Martin

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