The convergence of $\sum_{n=1}^\infty\frac{\mu(n+1)}{n+1}H_{n+1}$, where $\mu(n)$ is the Möbius function and $H_n=1+1/2+\ldots+1/n$

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.

• I hope that my calculations were rights, thanks all user by the patience and attention. – user243301 Nov 24 '16 at 21:02
• I'm confident that PNT in a quantitative form with relative error $o(1/\log^2n)$ together with partial summation yields the convergence of $\sum\mu(n)H_n/n$. – punctured dusk Nov 24 '16 at 21:15
• Many thanks for your contribution and attention @barto – user243301 Nov 24 '16 at 21:23

$$\sum_{n=1}^\infty\frac{\mu(n)\log n}n=-1.$$
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.