Let $M(N) := \sum_{n=1}^N \mu(n)$

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N$ for say $ K\ge2 $ is also sufficient. $K = 1$ is excluded as Merten's conjecture is false.)

If you show that for each $\epsilon > 0$ there are infinitely many values N such that $M(N) > N^{1-\epsilon}$.

$\textbf{Question 1}$: Does it invalidates R.H.?

$\textbf{Question 2}$: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ stand with R.H.?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H give sufficient conditions and discuss chains of necessity leading to R.H. But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

$\textbf{Notational remark: } $ Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes ( because LaTeX point is too small).

Let $M(N) := \sum_{n=1}^N \mu(n)$

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N$ for say $ K\ge2 $ is also sufficient. $K = 1$ is excluded as Mertens's conjecture is false.

Suppose you show that for each $\epsilon > 0$ there are infinitely many values N such that $M(N) > N^{1-\epsilon}$.

Question 1: Does this invalidate R.H.?

Question 2: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ stand with R.H.?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H. give sufficient conditions and discuss chains of necessity leading to R.H., But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

Notational remark: Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes (because LaTeX point is too small).

Let $M(N) := \sum_{n=1}^N \mu(n)$)

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N$ for say $ K\ge2 $ is also sufficient. $K = 1$ is excluded as Merten's conjecture is false.)

If you show that for each $\epsilon > 0$ there are infinitely many values N such that $M(N) > N^{1-\epsilon)}$.

$\textbf{Question 1}$: Does it invalidates R.H.?

$\textbf{Question 2}$: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ : stands with R.H.?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H give sufficient conditions and discuss chains of necessity leading to R.H. But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

$\textbf{Notational remark: } $ Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes ( because latex point is too small).

Let $M(N) := \sum_{n=1}^N \mu(n)$

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N$ for say $ K\ge2 $ is also sufficient. $K = 1$ is excluded as Merten's conjecture is false.)

If you show that for each $\epsilon > 0$ there are infinitely many values N such that $M(N) > N^{1-\epsilon}$.

$\textbf{Question 1}$: Does it invalidates R.H.?

$\textbf{Question 2}$: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ stand with R.H.?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H give sufficient conditions and discuss chains of necessity leading to R.H. But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

$\textbf{Notational remark: } $ Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes ( because LaTeX point is too small).

Let $M(N) := \sum_{n=1}^N \mu(n)$)

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N $ for say $ K>=2 $ is also sufficient. $K = 1$ is excluded as Merten's conjecture is false.)

If you show that for each $\epsilon > 0 $ there are infinitely many values N such that $M(N) > N^{1-\epsilon)}$  .

$\textbf{Question 1}$: Does it invalidates R.H.?

$\textbf{Question 2}$: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ : stands with R.H.  ?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H give sufficient conditions and discuss chains of necessity leading to R.H. But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

$\textbf{Notational remark: } $ Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes ( because latex point is too small).

Let $M(N) := \sum_{n=1}^N \mu(n)$)

It is known that bounding $M(N)$ by $N^{1/2+\epsilon}$ implies R.H.

A bound of $M(N)$ by $K\sqrt N$ for say $ K\ge2 $ is also sufficient. $K = 1$ is excluded as Merten's conjecture is false.)

If you show that for each $\epsilon > 0$ there are infinitely many values N such that $M(N) > N^{1-\epsilon)}$.

$\textbf{Question 1}$: Does it invalidates R.H.?

$\textbf{Question 2}$: How does a result like $ M(N)>N^{0,99} $ for any $N>N_o$ for some $N_o$ : stands with R.H.?

In the second question does it kill some well accepted heuristics about R.H.?

Most texts about number theory or R.H give sufficient conditions and discuss chains of necessity leading to R.H. But I found little about falsity of R.H. (the "heretic side" as called somewhere in Tao's blog).

$\textbf{Notational remark: } $ Mathematicians should use $\dot \mu(N)$ instead of $M(N)$, saving symbols and brain but not eyes ( because latex point is too small).