Let $\mu(n)$ the Möbius function, see the definition from this MathWorld. We consider the series $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)\mu(k)}{n^k k^n}.\tag{1}$$ I would like to know how to prove that this series is convergent.
My attempt was to use absolute convergence, but it fails: $$ \left|\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)\mu(k)}{n^k k^n} \right|\leq \sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^k k^n},$$ and $\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^k k^n}$ is divergent, since (this last series has positive terms) $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^k k^n}=\left(\sum_{n=1}+\sum_{n=2}^\infty\right)\sum_{k=1}^\infty\frac{1}{n^k k^n}=\sum_{k=1}^\infty\frac{1}{ k}+\sum_{n=2}^\infty\sum_{k=1}^\infty\frac{1}{n^k k^n}.$$
As motivation, but it is mainly curiosity, I've created $(1)$ when while reading [1] and I tried to set variations of those identities, calculating approximations with Wolfram Alpha, of series like than previous, or $$\sum_{k=1}^\infty\sum_{j=1}^k\frac{\mu(j)}{kj(k+j)}$$ or $$\sum_{k=1}^\infty\sum_{k=1}^k\frac{\mu(k)\mu(j)}{kj(k+j)}.$$ I don't know if such series were in the literature, or are interestings.
Question. Can you prove that $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(n)\mu(k)}{n^k k^n}$$ does converge? Many thanks.
One can see my approximation with Wolfram Alpha online calculator if type the code
sum mu(n)mu(k)/(k^n n^k), n=1 to 30, from k=1 to 30
or
sum mu(n)mu(k)/(k^n n^k), n=1 to 100, from k=1 to 100.
References:
[1] Whalter Janous, Around Apéry's Constant, Journal of Inequalities in Pure and Applied Mathematics, volume 7, issue 1, article 35 (2006).