# Prove that $\sum_{n=1}^\infty\frac{\mu(n)}{n}H_n\sum_{k=n+1}^\infty\frac{\mu(k)}{k^2}$ is convergent

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 , 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:

 Furdui, PROBLEMA 94 from La Gaceta de la Real Sociedad Matemática Española Vol. 11, No. 4 (2008).

• It's hard to immediately prove because of the conditional nature of convergence, but it seems highly likely. $\sum\frac{\mu(n)\ln n}{n}$ converges (to -1) and it's hard to imagine that the inner terms would 'conspire' to push the series towards divergence. – Steven Stadnicki Jul 1 '17 at 23:27
• I've no a good idea of how to study the convergence of this series. I know that it is required to study it using absolute convergence, and that we will need some asympotics. But I don't know a good strategy to solve the problem. – user243301 Jul 1 '17 at 23:27
• Many thanks @StevenStadnicki I was writting previous comment for all users. Thanks for your contribution. – user243301 Jul 1 '17 at 23:28
• $\sum_{n\geq 1}\frac{H_n}{n^2}$ is convergent and $\sum_{k\geq n+1}\frac{\mu(k)}{k^2}$ is bounded by $\frac{1}{n}$ in absolute value, so the given series is convergent. – Jack D'Aurizio Jul 1 '17 at 23:30
• Many thanks feel free to add you comment as hints for an answer @JackD'Aurizio And my apologizes to you and all users because seems a bad question. – user243301 Jul 1 '17 at 23:38

## 1 Answer

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.