Disclaimer: This could be a stupid question and has a very simple answer which I am unable to see.

Let $\{x_n\}_{n=1}^\infty$ be a sequence of vectors in a Hilbert space $$l^2_{k^{-2}}:=\{z=\{z(k)\}_{k=1}^\infty\in l^2:\sum\limits_{k=1}^\infty z(k)^2k^{-2}<\infty\}.$$ It is known that for some $x\in l^2_{k^{-2}}$ $$x_n(k)-x(k)=k(x_n(1)-x(1)), \ \ \forall k=1,2,\dots,n-1. \ \ \ \dots(A)$$ If we define $y\in l^2_{k^{-2}}$ such that $y(n)=x(n-1)-x(n); \ x(0)=0$ then it is also known that, $$x_n(1)-x(1)=\sum_{r\leq n}\frac{(\mu*y)(r)}r=\sum_{r=1}^n\frac{1}{r}\sum_{d|r}\mu\left(\frac{r}{d}\right)y(d)=\sum_{ab\leq n}\frac{\mu(a)}{a}\frac{y(b)}{b}.\dots(B)$$ where $\mu$ is the Mobius inversion function and $*$ is the Dirichlet convulution. From above its not difficult to see that $\lim\limits_{n\to\infty}x_n(1)=x(1)$ and hence for a fixed $k$; $\lim\limits_{n\to\infty}x_n(k)=x(k),$ from $(A)$ and $(B)$.

My question is from here how one can show that ,$$\|x_n-x\|_{l^2_{k^{-2}}}=\sum_{k=1}^\infty\frac{(x_n(k)-x(k))^2}{k^2}\xrightarrow{n\to\infty} 0?$$Using the information above it can be seen that the convergence highly depends on a good bound of $\sum\limits_{n\leq x}\frac{\mu(n)}{n}$. Even if assuming Riemann Hypothesis I am being unable to prove this.

ANY help is highly appreciated.