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I would like to prove the identity $$\sum_{\substack{b,d>0 \\ (b,d)=1}}\frac{1}{b^n}\frac{1}{d^m}=\frac{\zeta(n)\zeta(m)}{\zeta(m+n)},$$ where $\zeta$ is the Riemann zeta function and $n,m\ge 2$. Any help would be welcome.

El Brujo Especial
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Every pair $(r,s)$ of positive integers has the form $(tb,td)$ where $t=\gcd(r,s)$ and $\gcd(b,d)=1$. Therefore $$\zeta(m)\zeta(n)=\sum_{r,s>0}\frac1{r^n}\frac1{s^m} =\sum_{b,d>0\atop\gcd(b,d)=1}\sum_{t=1}^\infty\frac1{(tb)^n}\frac1{(td)^m} =\zeta(m+n)\sum_{b,d>0\atop\gcd(b,d)=1}\frac1{b^n}\frac1{d^m}.$$

Angina Seng
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  • Thank you! There is a typo in the exponents, I couldn't edit it though. – El Brujo Especial Jul 12 '20 at 21:01
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    Another method is to note that $1_{(b,d) = 1} = \sum_{c \mid (b,d)} \mu(c)$, so the original sum is $\sum_{b \geq 1} \sum_{d \geq 1} \sum_{c \mid (b,d)} \frac{\mu(c)}{b^n d^m}$. Now interchange the order of summation, so that this becomes $$\sum_{c \geq 1} \mu(c) \sum_{\substack{b \geq 1 \ b \equiv 0 \pmod{c}}} \frac{1}{b^n} \sum_{\substack{d \geq 1 \ d \equiv 0 \pmod{c}}} \frac{1}{d^m}.$$ The result then follows upon making the change of variables $b \mapsto cb$ and $d \mapsto cd$. – Peter Humphries Jul 12 '20 at 22:33