I am trying to prove
$$ \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{\mathrm{lcm}(n,m)}{\mathrm{max}(n,m)^5} = \frac{\pi^6}{540\zeta(3)},$$
my guess is that the solution uses Euler products or Dirichlet series.
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2$\begingroup$ Use lcm times gcd is product and with $k$ the gcd the sum in $k$ is $\zeta^4$ while the remaining is just $\sum_{b \leq a, (a,b)=1}\frac{b}{a^4}$ and that is standard stuff ( eg sum on $b$ gives $a\phi(a)/2$ etc $\endgroup$Conrad– Conrad2026-02-28 16:49:26 +00:00Commented 9 hours ago
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as @Conrad pointed it out , let $\gcd(n,m)=d$, so $n=ad$ and $m=bd$ where $\gcd(a,b)=1$
and we have $\displaystyle \text{lcm}(n,m)=\frac{nm}{\gcd(n,m)}=abd$ and $\max(n,m)=d \max(a,b)$
so $S=\displaystyle \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{\mathrm{lcm}(n,m)}{\mathrm{max}(n,m)^5} = \sum_{d=1}^{\infty}\sum_{\quad a,b \ge 1 \\ \gcd(a,b)=1} \frac{abd}{d^5(\max(a,b))^5}= \sum_{d=1}^{\infty} \left(\frac{1}{d^4} \sum_{\quad a,b \ge 1 \\ \gcd(a,b)=1}\frac{ab}{\max(a,b)^5} \right)$
$\displaystyle \Rightarrow S= \zeta(4) \sum_{\quad a,b \ge 1 \\ \gcd(a,b)=1}\frac{ab}{\max(a,b)^5}=\zeta(4)(\sum_{ \quad a\ge b \ge 1 \\ \gcd(a,b)=1}\frac{ab}{a^5}+\sum_{ \quad b\ge a \ge 1 \\ \gcd(a,b)=1}\frac{ab}{b^5})$
$\displaystyle \Rightarrow S= 2\zeta(4)\sum_{\quad a \ge b \ge 1 \\ \gcd(a,b)=1} \frac{b}{a^4} = 2\zeta(4) \sum_{a=1}^{\infty}\frac{1}{a^4}\sum_{\quad a \ge b \ge 1 \\ \gcd(a,b)=1}b$
we now use this sum identity, $\displaystyle\sum_{\quad a \ge b \ge 1 \\ \gcd(a,b)=1}b=\frac{a \varphi(a)}{2}$ so $\displaystyle S=\zeta(4)\sum_{a=1}^{\infty}\frac{\varphi(a)}{a^3}$
and $\displaystyle \sum_{n=1}^{\infty} \frac{\varphi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}$ so $\displaystyle S=\zeta(4) \frac{\zeta(2)}{\zeta(3)}= \frac{\pi^6}{540 \zeta(3)}$
