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In the first page of this document by Sitaramachandrarao it is written that $$\sum_{n\leq x} \frac{1}{\phi (n)} = A( \log x + B) + E_0 (x) $$ and $$ \sum_{n\leq x} \frac{n}{\phi (n)} = Ax - \frac{1}{2} \log x + E_1 (x),$$ where $$E_0 (x) = O\left( \frac{\log x}{x} \right) , A = \frac{315 \zeta (3)}{2\pi ^4} \text{ and } B = \gamma - \sum_{p} \frac{\log p}{p^2 - p +1},$$ where $\zeta$ is the Riemann zeta function and $\gamma$ is the Euler-Mascheroni constant (I could not find what $E_1(x)$ denotes, so I think it denotes $O\left( \frac{\log x}{x} \right)$. To this end, is there a way to deduce that

$$\sum_{n \leq x} \frac{1}{\phi (n)} = C\log x + O(1), \qquad{} C = \sum \frac{1}{n \phi (n)}$$

where the last sum equal to $C$ ranges over all square-free integers? We know that $$\sum_{d|n}\frac{\mu^2(d)}{\phi(d)} =\frac{n}{\phi(n)}$$ (I can attach a proof if needed) and also $$ \sum_{n\leq x}\frac{1}{\varphi(n)} = \sum_{n\leq x}\frac{1}{n}\cdot\frac{n}{\varphi(n)} = O\left(\sum_{n\leq x}\frac{1}{n}\right) = O(\log n).$$ I also found this question which says we can obtain the asymptotic formula $$\sum_{n \leq x} \frac{\mu^2(n)}{\varphi(n)} = \log x + c + o(1),$$ where $$c = \gamma + \sum_{p} \frac{\log p}{p(p-1)} = 1.332\ldots.$$

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  • $\begingroup$ I wonder what your questions are. It seems to me that your questions are answered in the paper by Sitaramachandrarao $\endgroup$ – i707107 Mar 9 '18 at 0:06
  • $\begingroup$ @i707107 Instead of $\frac{315 \zeta (3)}{2\pi ^4} \log x$ I wanted to see if we could achieve $ \left( \sum \frac{1}{n \phi (n)} \right) \log x$ $\endgroup$ – Compact Mar 9 '18 at 0:13
  • $\begingroup$ So, your question is "Why is $\frac{315\zeta(3)}{2\pi^4} = \sum \frac 1{n\phi(n)}$? $\endgroup$ – i707107 Mar 9 '18 at 0:34
  • $\begingroup$ Also, the sum on the right is restricted to square free numbers, so it is really $\sum \frac{\mu^2(n)}{n\phi(n)}$. $\endgroup$ – i707107 Mar 9 '18 at 0:35
  • $\begingroup$ @i707107 Yes, that is my question. $\endgroup$ – Compact Mar 9 '18 at 2:03
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Euler's product gives $$\sum_{n\geq 1}\frac{(\mu^2/\varphi)(n)}{n^s}=\prod_{p}\left(1+\frac{1}{(p-1)p^s}\right) $$ from which $$ \sum_{n\geq 1}\frac{(\mu^2/\varphi)(n)}{n}=\prod_{p}\frac{p^2-p+1}{p(p-1)}=\prod_p\frac{p^3+1}{p(p^2-1)}=\prod_{p}\frac{(p^6-1)}{p(p^2-1)(p^3-1)}=\frac{\zeta(2)\zeta(3)}{\zeta(6)}$$ and the RHS equals $\frac{315\,\zeta(3)}{2\pi^4}$ as wanted.

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  • $\begingroup$ Damn, Jack, you do nice stuff! $\endgroup$ – marty cohen Mar 9 '18 at 17:20

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