Let $0<y\leq 1$ a real fixed number and we denote by $\mu(n)$ the Möbius function, and with $\{x\}$ the fractional part function. I don't know if the follwing calculations were right, because I want to show convergence using the strict inequality $<$ deduced from the definition of the fractional part function but in calculations involving the harmonic series, that I know that diverges. See my attempt after the following:
Question. Let $0<y\leq 1$ a real fixed number, does converge $$\sum_{k=1}^\infty\frac{\mu(k)\{\frac{1}{ky}\}}{k}?$$ I asked myself this question after I was studying the convergence of some infinite products (isn't directly related, but if it is interesting for you see my next post). Many thanks.
By Abel's identity I write ($k\leq x$ instead of $1\leq k\leq x$ for a fixed real $x>1$) $$\sum_{k\leq x}\frac{\mu(k)\{\frac{1}{ky}\}}{k}=\frac{1}{x}\left(\sum_{k\leq x}\mu(k)\{\frac{1}{ky}\}\right)+\int_1^x\frac{\sum_{k\leq t}\mu(k)\{\frac{1}{ky}\}}{t^2}dt,$$ and I know that $\sum_{k=1}^\infty\mu(k)\{\frac{1}{ky}\}$ converges, because was in the literature, thus the first term in RHS tend to zero as $x\to\infty$.
After (and here will be the doubt) I want to show the convergence (if it is possible), well for LHS from RHS, or for RHS from LHS, that is the same. My way was use that the fractional part satisfies as function $0\leq\{X\}<1$ then I wrote $$\left|\int_1^x\right|=\left|\sum_{k\leq x}\frac{\mu(k)\{\frac{1}{ky}\}}{k}\right|\leq\sum_{k\leq x}\frac{1\cdot\{\frac{1}{ky}\}}{k}<\sum_{k\leq x}\frac{1}{k},$$
the second inequality is where I believe that could be problems when I say: when $x\to\infty$ the integral/series converges absolutely since one has the strict inequality $<\infty$. It is right? Can you explain if it is right or give a different reasoning to discuss the convergence of such infinite series in the Question?
For sufficiently large $k$, $$\left\{\frac1{ky}\right\}=\frac1{ky}.$$
Then $$ \sum_{k=1}^{\infty}\frac{\left\{\frac1{ky}\right\}}{k} $$ converges since $$ \sum_{k=1}^{\infty} \frac1{k^2 y} $$ converges.
Therefore, by $|\mu(k)|\leq 1$, your series converges absolutely.