Write the sum $\sum\limits_{a \in \mathbb{N}}\sum\limits_{b \in \mathbb{N}} \frac{(a,b)}{a^sb^t}$ in terms of the Riemann zeta function

Question

I have the following exercise, and I need some help:

Write the sum $$\sum\limits_{a \in \mathbb{N}}\sum\limits_{b \in \mathbb{N}} \frac{(a,b)}{a^sb^t}$$ in terms of the Riemann zeta function ($(a,b)$ is the greatest common divisor). I assume you have to somehow sort the sums for $d:=(a,b)$, but I have no idea how, any help is appreciated.

I tried to use the fact that $d=xa + yb$ and then $$ \frac{(a,b)}{a^sb^t} = \frac{x}{a^{s-1}b^t} + \frac{y}{a^sb^{t-1}}$$ but since I don't know if the original series is convergent or not, I can't use re-arrangements and so I got stuck.

Answer 1

First, define

$$f(s,t;u)=\sum_{a=1}^\infty\sum_{b= 1}^\infty\frac{(a,b)^u}{a^sb^t}$$

We seek to derive a formula for $f(s,t;1)$, though this process will naturally lead us toward a more general formula for $f(s,t;u)$. We may perform the following manipulations:

\begin{equation} \begin{split} f(s,t;u)&=\sum_{a=1}^\infty\sum_{b=1}^\infty\frac{(a,b)^u}{a^sb^t}\\ &=\sum_{r=1}^\infty\sum_{(a,b)=r}\frac{r^u}{a^sb^t}\\ &=\sum_{r=1}^\infty\sum_{(a,b)=1}\frac{r^u}{(ar)^s(br)^t}\\ &=\sum_{r=1}^\infty\frac{1}{r^{s+t-u}}\sum_{(a,b)=1}\frac{1}{{a}^s{b}^t}\\ &=\zeta(s+t-u)\sum_{(a,b)=1}\frac{1}{a^sb^t}\\ \end{split} \end{equation}

Now, simply note that $f(s,t;0)=\zeta(s)\zeta(t)$, so setting $u=0$ and rearanging the above equation, we have that

$$\sum_{(a,b)=1}\frac{1}{a^sb^t} =\frac{f(s,t;0)}{\zeta(s+t)}=\frac{\zeta(s)\zeta(t)}{\zeta(s+t)}$$

and therefore,

$$\sum_{a=1}^\infty\sum_{b= 1}^\infty\frac{(a,b)^u}{a^sb^t}= f(s,t;u)=\zeta(s+t-u)\sum_{(a,b)=1}\frac{1}{a^sb^t}=\frac{\zeta(s+t-u)\zeta(s)\zeta(t)}{\zeta(s+t)}$$

Finally, setting $u=1$, we have your particular case:

$$\sum_{a=1}^\infty\sum_{b= 1}^\infty\frac{(a,b)}{a^sb^t}=\frac{\zeta(s+t-1)\zeta(s)\zeta(t)}{\zeta(s+t)}$$

Answer 2

Using the convolution property of Euler's totient function, we have

\begin{aligned} \sum_{a\ge1}\sum_{b\ge1}{(a,b)\over a^sb^t} &=\sum_{a\ge1}\sum_{b\ge1}{1\over a^sb^t}\sum_{d|(a,b)}\varphi(d)=\sum_{d\ge1}\varphi(d)\sum_{\substack{a\ge1\\d|a}}{1\over a^s}\sum_{\substack{b\ge1\\d|b}}{1\over b^t} \\ &=\sum_{d\ge1}{\varphi(d)\over d^{s+t}}\sum_{\substack{a\ge1\\d|a}}{1\over(a/d)^s}\sum_{\substack{b\ge1\\d|b}}{1\over(b/d)^t}=\sum_{d\ge1}{\varphi(d)\over d^{s+t}}\sum_{u\ge1}{1\over u^s}\sum_{v\ge1}{1\over v^t} \\ &={\zeta(s+t-1)\over\zeta(s+t)}\cdot\zeta(s)\zeta(t). \end{aligned}