For integers $n\geq 1$ let $$H_n=1+\frac{1}{2}+\ldots+\frac{1}{n}$$ the $nth$ harmonic number, and $\mu(n)$ the Möbius function. See, if you need it, this Wikipedia to know the definition of Möbius function.
After I've read a problem due to Furdui from [1], I would like to ask about the convergence of a similar series:
Question. What is a good way to deduce if the series $$\sum_{n=1}^\infty\frac{\mu(n)}{n}H_n\sum_{k=n+1}^\infty\frac{\mu(k)}{k^2}$$ does converge? Thanks in advance.
References:
[1] Furdui, PROBLEMA 94 from La Gaceta de la Real Sociedad Matemática Española Vol. 11, No. 4 (2008).
With a very crude estimation $$\left|\sum_{k\geq n+1}\frac{\mu(k)}{k^2}\right|\leq \sum_{k\geq n+1}\frac{1}{k^2}\leq \int_{n}^{+\infty}\frac{dx}{x^2}=\frac{1}{n} $$ since $\left|\mu(k)\right|\leq 1$. Since $$\sum_{n\geq 1}\frac{H_n}{n^2}=2\,\zeta(3) $$ is a convergent series, the given series is convergent too.