I was looking for information about the Riemann Hypothesis, and I found that the Riemann Hypothesis is equivalent to the statement $$\sum_{k\le n}\frac{\mu(k)}{k}=O(n^{-1/2+\varepsilon})$$
If we consider the sum $S=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}$ and we expand it, we have that $$S=1-\frac{1}{2}-\frac{1}{3}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}+\frac{1}{10}-\frac{1}{11}-\frac{1}{13}+...$$
Where the denominators are all the square free integers.
I noticed that if we set $$\frac{1}{2}S=\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{10}+\frac{1}{12}-\frac{1}{14}+\frac{1}{20}-\frac{1}{22}-\frac{1}{26}+...$$
And we subtract, most of the terms vanish and we get that $$\left(1-\frac{1}{2}\right)S=\frac{1}{4}+\frac{1}{12}+\frac{1}{20}+\frac{1}{28}+\frac{1}{44}+...$$
$$\frac{1}{2}S=\frac{1}{4}\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+...\right)$$
$$\frac{1}{2}S=\frac{1}{8}+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+...\right)$$
And finally, $$S=\frac{1}{4}+\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+...\right)$$
Therefore, it seems we can establish that $$\sum_{k=1}^{\infty}\frac{\mu(k)}{k}=\frac{1}{4}+\frac{1}{2}\sum_{p}\left(\frac{1}{p}\right)$$
As partial sums of $\sum_{p}\left(\frac{1}{p}\right)$ are well known by Mertens' second theorem, my question is: is my reasoning valid? could it help or give some additional information to get the desired statement $\sum_{k\le n}\frac{\mu(k)}{k}=O(n^{-1/2+\varepsilon})$?