Here in Wikipedia we've the identity for harmonic numbers $$\sum_{k=1}^n H_k=(n+1)(H_{n+1}-1)$$ (I believe that if there are convergence in each summand you can experiment with different identities involving harmonic numbers following my approach, but I don't use any property of harmonic numbers, only previous identity) then we multiply by $\frac{\mu(n+1)}{(n+1)^2}$ for integers $n\geq 1$, and we take the sum $$\sum_{n=1}^\infty\frac{\mu(n+1)}{(n+1)^2}\sum_{k=1}^n H_k=\sum_{n=1}^\infty\frac{\mu(n+1)}{n+1}H_{n+1}-\sum_{n=1}^\infty\frac{\mu(n+1)}{n+1}.$$ Notice that if we presume convergence then using the Prime Number Theorem $$\sum_{n=1}^\infty\frac{\mu(n+1)}{n+1}=\left(\sum_{n=1}^\infty\frac{\mu(n)}{n}\right)-1=0-1=-1,$$ thus if we presume convergence for each summand ($\sum_{n=1}^\infty=\lim_{N\to\infty}\sum_{n=1}^N$) one can write $$\sum_{n=1}^\infty\frac{\mu(n+1)}{(n+1)^2}\sum_{k=1}^n H_k=1+\sum_{n=1}^\infty\frac{\mu(n+1)}{n+1}H_{n+1}.$$
Question. It's known or can you deduce the convergence for some of these summands $$\sum_{n=1}^\infty\frac{\mu(n+1)}{(n+1)^2}\sum_{k=1}^n H_k=1+\sum_{n=1}^\infty\frac{\mu(n+1)}{n+1}H_{n+1}?$$ Since I would like to know if the series in the title of this post is convergent, you can work with such or well with the series in LHS. Many thanks.
My attempt was using absolute convergence and I've calculated with Wolfram Alpha some partial sums, change your $N$ instead my $1000$ in the code in the box.
As it turns out,
$$\sum_{n=1}^\infty\frac{\mu(n)\log n}n=-1.$$
(Found as Exercise 16, chapter 6 in H. Montgomery and C. Vaughan, Multiplicative Number Theory: I. Classical Theory, Cambridge studies in advanced mathematics, 2006, which on its turn references to work of Landau.)
Using $H_n=\log n+\gamma+O(1/n)$ we get $$\sum_{n=1}^\infty\frac{\mu(n)H_n}n=-1+\sum_{n=1}^\infty\frac{\mu(n)(H_n-\log n-\gamma)}n$$ and the latter series converges absolutely.