2
$\begingroup$

I would like to show that as X approaches infinity,

$$\sum_{n\leq X} \frac{1}{\phi(n)} \sim \log(X)\cdot\sum_{k=1}^{\infty} \frac{\mu(k)^2}{k\cdot\phi(k)}.$$

I have already proven

$$\sum_{n\leq X} \frac{1}{\phi(n)} = \sum_{k\leq X} \left[\frac{\mu(k)^2}{k\cdot\phi(k)}\cdot\sum_{t\leq X/k} \frac1t\right].$$

I recognize that $\Sigma_{t\leq X/k} 1/k$ is the $X/k$th harmonic number and that harmonic numbers can be approximated by $\ln(X/k)$.
Wolfram Alpha says that $$\sum_{k=1}^{\infty} \frac{\mu(k)^2}{k\cdot\phi(k)} = \frac{315\zeta(3)}{2\pi^4}.$$

I have a feeling an epsilon-delta proof might be appropriate here, but I'm not sure where to proceed.

  • 0
    can you edit \leqX to \leq X and \infinity to \infty2011-04-01
  • 0
    You seem to be misunderstanding how the braces around subscripts and superscripts work. Braces serve to group things together into a single unit. Thus "a_{b^c}" yields $a_{b^c}$, not $a_b^c$, whereas "a_b^\infty" ($a_b^\infty$) is OK without braces, since \infty is already a single unit.2011-04-01
  • 0
    They should all be fixed now.2011-04-01
  • 0
    By $\sum_{t \le X/k} 1/k$ do you mean $\sum_{t \le X/k} 1/t$?2011-04-01

1 Answers 1

1

Found an answer. $\sum_{t\leq X/k} 1/t = \log(X/k) + \gamma + O(1/(X/k))$ When you multiply it out, the sum can be split into four terms:

$$ \sum_{n\leq X} 1/\phi(n) = \log(X)*\sum_{k\leq X} \mu(k)^2/(k*\phi(k)) - \sum_{k\leq X} \mu(k)^2*\log(k)/(k*\phi(k)) $$

$$ + \sum_{k\leq X} \gamma*\mu(k)^2/(k*\phi(k)) + \sum_{k\leq X} \mu(k)^2/(k*\phi(k)*O(k/X)) $$

I had also previously shown that $\sum_{k\leq X} \mu(k)^2/(k*\phi(k))$ and $\sum_{k\leq X} \mu(k)^2*\log(k)/(k*\phi(k))$ converge, so as X approaches infinity, the last three terms of the sum become constants, leaving only the first term. Thus we can conclude that

$$ \lim_{X\to \infty} \frac{\sum_{n\leq X} 1/\phi(n)}{\sum_{k\leq X} \mu(k)^2/(k*\phi(k))} = 1 $$