Summation function of the inverse of Euler function

Question

How can I show that $$ \sum_{n \leq x} \frac{1}{ \varphi(n)}= \frac{ \zeta(2) \zeta(3)}{ \zeta(6)}\log(x)+O(1)?$$

Answer 1

Its Shown in this Wikipedia page about Euler's phi function : Euler's totient function

$\sum \limits_{k=1}^{x} \frac{1}{\phi(k)} = \frac{315 \zeta(3)}{2 \pi^4}(\ln(x)+\gamma+o(1))$ actually even stronger result than $o(1)$ but thats you are asking about.

also the proving method is shown in this paper : On an error term of Landau II

hope you found what you are looking for.

Answer 2

An elementary proof is as follows:

We use the following: $$ \frac1{\phi(n)}=\sum_{d|n} \frac 1d \frac{\mu(\frac nd)^2}{\frac nd \phi(\frac nd)}.$$

This identity follows from first proving $$ \frac n{\phi(n)} = \sum_{d|n} \frac{\mu(d)^2}{\phi(d)}. $$ Then we have $$ \begin{align} \sum_{n\leq x } \frac1{\phi(n)} &= \sum_{n\leq x}\sum_{d|n} \frac 1d \frac{\mu(\frac nd)^2}{\frac nd \phi(\frac nd)} =\sum_{d\leq x}\frac 1d \sum_{k\leq \frac xd} \frac{\mu(k)^2}{k\phi(k)}\\ &=\sum_{k\leq x} \frac{\mu(k)^2}{k\phi(k)} \sum_{d\leq \frac xk} \frac 1d=\sum_{k\leq x} \frac{\mu(k)^2}{k\phi(k)} \left(\log \frac xk + O(1)\right)\\ &=\sum_{k=1}^{\infty} \frac{\mu(k)^2}{k\phi(k)} \log x + O(1). \end{align} $$ It can be also shown elementary way (use Euler product) that $$ \sum_{k=1}^{\infty} \frac{\mu(k)^2}{k\phi(k)}=\frac{\zeta(2)\zeta(3)}{\zeta(6)} $$