Summation function of the inverse of Euler function How can I show that $$  \sum_{n \leq x} \frac{1}{ \varphi(n)}= \frac{ \zeta(2)  \zeta(3)}{ \zeta(6)}\log(x)+O(1)?$$
 A: 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 
$$
\sum_{k=1}^{\infty}  \frac{\mu(k)^2}{k\phi(k)}=\frac{\zeta(2)\zeta(3)}{\zeta(6)}
$$
A: 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.
