# On the convergence of $\sum \mu(n)/n^s$

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. :)

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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.
Thanks David. I followed on similar lines as by Beurling. Basically, I showed, that this expression satisfies: $$\left| \frac{\eta(s)}{s}\sum_{k=1}^{\infty} \frac{\mu(n)}{n^s} - \frac{1 - 2^{1-s}}{s} \right| \leq || 1 + f_\mu||_2 . || x^{s-1} ||_2$$ where $f_\mu(1/x) = \sum \mu(n)\nu(1/nx)$, $\nu(n) = \rho(x/2) + 1/2 - \rho(x/2 + 1/2)$ where $\rho(x)$ is the fractional part of $x$, and the norm is on $L^2(0,1/2)$ . Then I proceeded to show that $$\lim_{n\to\infty} \left( \int_{0}^{1/2} \left|1 + \sum_{k=1}^{n} \mu(k) \nu(1/kx)\right|^2 dx \right)^{1/2} < \infty$$ – Roupam Ghosh Feb 10 '11 at 23:11
That part is ready, and it does form a Cauchy sequence and this time the result shows that the expression in the main question is zero for $\Re(s) > 1/2$. :) ( I am saying this with a pessimistic voice since the implications are huge and hence the chances of having a unseen bug in my calculations are high :P ) – Roupam Ghosh Feb 11 '11 at 7:50