# 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.

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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
But my point is, how can you make ANY statement which involves this sum, if you have not first proved that this sum converges? What does your statement MEAN if you haven't proved that? –  David Speyer Feb 11 '11 at 0:58
Thanks, for specifying what is "still needed" to show. I have prepared my paper, but I followed along the lines of "New versions of the Nyman-Beurling criterion for the Riemann Hypothesis" by Baez-Duarte. I think I will attach a section proving that it does form a Cauchy sequence. I will comment here once that part is ready. –  Roupam Ghosh Feb 11 '11 at 2:54
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
Nope... Its got a flaw... and so I am not going to post it... :P –  Roupam Ghosh Feb 13 '11 at 0:54