On the error term of an aproximation to $\sum_{k\le n}\frac{\mu(k)}{k}$ through Legendre's formula I was looking for information about the Riemann Hypothesis, and I found that the Riemann Hypothesis is equivalent to the statement $$\sum_{k\le n}\frac{\mu(k)}{k}=O(n^{-1/2+\varepsilon})$$
The well known Legendre's formula states that $$\pi(n)-\pi\left(\sqrt{n}\right)+1=\lfloor n\rfloor-\sum_{p_{i}\leq\sqrt{n}}\lfloor\frac{n}{p_{i}}\rfloor+\sum_{p_{i}<p_{j}\leq\sqrt{n}}\lfloor\frac{n}{p_{i}p_{j}}\rfloor-...$$
Therefore, if the error term between $\lfloor n\rfloor-\sum_{p_{i}\leq\sqrt{n}}\lfloor\frac{n}{p_{i}}\rfloor+\sum_{p_{i}<p_{j}\leq\sqrt{n}}\lfloor\frac{n}{p_{i}p_{j}}\rfloor-...$ and $ n-\sum_{p_{i}\leq\sqrt{n}}\frac{n}{p_{i}}+\sum_{p_{i}<p_{j}\leq\sqrt{n}}\frac{n}{p_{i}p_{j}}-...$ is small (which it seems), we could affirm roughly applying the Prime Number Theorem that $$n\sum_{k\le n}\frac{\mu(k)}{k}\approx\frac{n}{\log n}-\frac{\sqrt{n}}{\log\left({\sqrt{n}}\right)}+1+E(x)$$
Therefore, $$\sum_{k\le n}\frac{\mu(k)}{k}\approx\frac{1}{\log n}-\frac{2\sqrt{n}}{n\log{n}}+\frac{1}{n}+\frac{E(x)}{n}$$
$$\sum_{k\le n}\frac{\mu(k)}{k}\approx\frac{n-2\sqrt{n}+\log{n}+E(x)\log n}{n\log n}<\frac{1}{\log{n}}+\frac{E(x)}{n}$$
As a consecuence, if the reasoning is good, I find that bounding the mentioned error term would be a possible approach for binding $\sum_{k\le n}\frac{\mu(k)}{k}$. Is there any available information on the research progress on this error term binding?
Thanks in advance!
 A: The devil is in this detail:
"Therefore, if the error term between $\lfloor n\rfloor-\sum_{p_{i}\leq\sqrt{n}}\lfloor\frac{n}{p_{i}}\rfloor+\sum_{p_{i}<p_{j}\leq\sqrt{n}}\lfloor\frac{n}{p_{i}p_{j}}\rfloor-...$ and $ n-\sum_{p_{i}\leq\sqrt{n}}\frac{n}{p_{i}}+\sum_{p_{i}<p_{j}\leq\sqrt{n}}\frac{n}{p_{i}p_{j}}-...$ is small (which it seems)...."
It's true of course that each individual error, when replacing $\lfloor\cdot\rfloor$ by $\cdot$, is small (less than $1$ in absolute value). The problem is that there are an enormous number of errors: if $P=\pi(\sqrt x)$ is the number of primes up to $\sqrt n$, then there are $P$ errors in the first sum, $\binom P2$ errors in the second sum, $\binom P3$ errors in the third sum—and note that $\binom P3$ is already a larger magnitude than $n^{1/49}$ say, which is far larger than the main term. (The total number of error terms is $2^P$ which is even larger.)
This is the reason that the "sieve of Eratosthenes" is not sufficiently strong to get an asymptotic formula for the number of primes up to $n$.
