Timeline for Bounding a series involving Ramanujan's sum
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2017 at 7:57 | comment | added | The Number Theorist | @GregMartin It's true! I agree with you! In fact, I'm trying to use this method. | |
| Mar 31, 2017 at 23:56 | comment | added | Greg Martin | Note that the Dirichlet series $\sum_{\ell=1}^\infty \frac{\mu(\ell)\ell^2}{\phi(\ell)^2} \ell^{-s}$ is $\frac1{\zeta(s)^2}$ times a series that converges absolutely down to $\Re(s)>\frac12$. We want the tail of this series at $s=2$. It should be possible, using contour integration a la the proof of the PNT, to get a bound $\sum_{\ell>N} \frac{\mu(\ell)}{\phi(\ell)^2} \ll \frac1N \exp(-\sqrt{c\log N})$ for a small $c>0$. | |
| Mar 31, 2017 at 21:41 | comment | added | The Number Theorist | @GregMartin In fact one can prove that $\sum_{l>r}\frac{\mu^{2}(l)}{\varphi^{2}(l)}\leq \frac{6.7345}{r}$. | |
| Mar 31, 2017 at 21:39 | comment | added | The Number Theorist | Thanks for your answer! ..but this is not what I was looking for. In fact, you give an estimate for the sum of the absolute values of the general term, and this is easy, as you showed. I yet known the bound with $\frac{\tau(n)n}{\varphi(n)}$, but this is to much big for my purposes. As I said above, I want a bound on the sum itself, that probably is of the form $\frac{C}{r}\frac{n}{\varphi(n)}$. Using your method, one reduce to find a bound for $\sum_{l>\frac{r}{d}}\frac{\mu(l)}{\varphi(l)^{2}}$. | |
| Mar 31, 2017 at 16:51 | comment | added | Greg Martin | From the general philosophy "$\phi(m)\asymp m$ on average", we should be able to get rid of the $\log r$ term. Indeed, $\sum_{\ell\le x} \frac{\ell^2}{\phi(\ell)^2}$ can be shown to be $O(x)$ by standard multiplicative-function methods, and then $\sum_{U\le\ell\le2U} \frac1{\phi(\ell)^2} \ll \frac1U$ follows easily and implies $\sum_{U\ge N} \frac1{\phi(\ell)^2} \ll \frac1N$. | |
| Mar 31, 2017 at 15:13 | history | answered | Matt Young | CC BY-SA 3.0 |