I arrived at something during my maths ponderings which is really exciting for me.
It is clearly stated in the book on Riemann Hypothesis by Borwein that the convergence of $\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$ for $\Re(s) > 1/2$ is necessary and sufficient for RH. This is ofcourse valid since, $\sum \mu(n)/n^s = 1/\zeta(s)$ for $\Re(s) > 1$
Having said that, I have reached a point where I got, for $\Re(s) > 1/2$ $$ \left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| < \infty $$ $\eta(s)$ is the Dirichlet eta function.
My question is,
What do I interpret out of this formula. Does this result imply RH, or falls short of it?
I believe the second option might be more correct, because this result does not say anything about the zeros of the eta function. But at least it is clear that if $\sum \mu(n)/n^s$ blows up then $\eta(s)$ also must have a zero to bring it down.
Any elaborated answer will be highly appreciated, because it is a current work in progress. :)
Let me make sure I understand what you mean by the formula $$ \left| \frac{\eta(s)}{s} \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1-2^{1-s}}{s} \right| < \infty. $$
Do you actually mean that you have proven that this sum is convergent for $s$ in the critical strip and that, for such $s$, this inequality holds? Because, if you have shown that this sum converges in that range, you have already shown RH.
I can think of other interpretations of your statement, but I will wait for clarification before elaborating on them. Here is a basic point to remember: The statement that $\sum_{n=1}^{\infty} a_n$ converges is a statement about the partial sums $\sum_{n=1}^N a_n$. (Namely, that they form a Cauchy sequence.) If you never say anything about these partial sums, it is unlikely you have proved convergence.