Is an asymptotic formula of $$\sum_{k=1}^n \frac{1}{\varphi(k)}$$ known ?
The infinite sum $$\sum_{k=1}^\infty \frac{1}{\varphi(k)}$$ diverges which can be shown by comparing it to the harmonic series $$\sum_{k=1}^\infty \frac{1}{k}$$
Numerical values :
? sum(j=1,10^7,1.0/eulerphi(j))
%44 = 31.26649923752769616067244698051873057166
? sum(j=1,10^8,1.0/eulerphi(j))
%45 = 35.74179524657529982203363898007376015266
?
So, the value seems to be roughly $1.94\cdot \ln(n)$. Is this actually true, and if yes, what is known about the constant near $1.94$ ?
Best Answer
From the Wikipedia article on the totient function, we have $$\sum_{k=1}^n\frac{1}{\varphi(k)}=\frac{315\zeta(3)}{2\pi^4}\left(\log n + \gamma -\sum_{p \in\mathbb{P}}\frac{\log p}{p^2-p+1}\right)+O\left(\frac{(\log n)^{\frac{2}{3}}}{n}\right),$$ so the proportionality constant you found was $\frac{315\zeta(3)}{2\pi^4}$. It was proved in a paper in 1900 by Landau according to this source.