Let $\mu(n)$ the Möbius function, then inspired in this Mathematics Stack Exchange post my question will be
Question. How do you prove that the sequence $$a_n:=\sum_{k=1}^{n-1}\mu(k)\left(\frac{n}{k^2(n-k)}\right)^{2},$$ defined for $n\geq 2$, does converge as $n\to\infty$? Many thanks.
I can state these calculations and reasonings:
0) I know the criterion for bounded and monotonic sequences, but I don't know how discuss the monotonicity of the sequence.
1) One has $a_n\leq n^2\sum_{k=1}^{n-1}\frac{\mu(k)}{k^4}$, (since $(n-k)^2\geq 1$ implies $1/(n-k)^2\leq 1$), but I am not sure if it could be combined with Stolz theorem succesfully (I believe that no). Additionally we know the triangle inequality, and $|\mu(k)|\leq 1$.
2) Since $\frac{1}{k^2}+\frac{1}{n-k}=\frac{n}{k^2(n-k)}-\frac{k}{k^2(n-k)}+\frac{k^2}{k^2(n-k)}$ implies $\frac{n}{k^2(n-k)}=\frac{1}{k^2}+\frac{1}{k(n-k)}$, and thus I've deduced $$a_n=\sum_{k=1}^{n-1}\mu(k)\left(\frac{1}{k^2}+\frac{1}{k(n-k)}\right)^2=\sum_{k=1}^{n-1}\frac{\mu(k)}{k^4}+\sum_{k=1}^{n-1}\frac{\mu(k)}{k^2(n-k)^2}+2\sum_{k=1}^{n-1}\frac{\mu(k)}{k^3(n-k)}.$$ And I know that the first term of $RHS$ is convergent, one has the well known $$\sum_{k=1}^\infty\frac{\mu(k)}{k^4}=\frac{1}{\zeta(4)}=\frac{90}{\pi^4}.$$ And I believe isn't possible to exploit the symmetry.
3) With Wolfram Alpha online calculator I was playing with codes like these
sum mu(k)(10000/(k^2(10000-k)))^2, from k=1 to 9999
sum mu(k)(20000/(k^2(20000-k)))^2, from k=1 to 19999
Hint: $$\sum_{k=1}^{n-1}\left|\frac{\mu\left(k\right)}{k^{2}}\left(\frac{1}{k}+\frac{1}{n-k}\right)^{2}\right|\leq\sum_{k=1}^{n-1}\frac{1}{k^{4}}+\sum_{k=1}^{n-1}\frac{1}{k^{2}\left(n-k\right)^{2}}+\frac{2}{n}\sum_{k=1}^{n-1}\frac{1}{k^{2}}\left(\frac{1}{k}+\frac{1}{n-k}\right)$$ $$\leq\sum_{k=1}^{n-1}\frac{1}{k^{4}}+\sum_{k=1}^{n-1}\frac{1}{k^{2}}+\frac{2}{n}\sum_{k=1}^{n-1}\frac{1}{k^{2}}\left(\frac{1}{k}+1\right).$$