It is well-known that in average $\varphi(n)$ behaves like $\frac1{\zeta(2)}n=\frac{6}{\pi^2}n$. But it looks that in some sense it is ``asymptotically larger''. In particular, the ratio 
$$
\zeta(2)(1-t)^2\sum\_{n=1}^{\infty} \varphi(n)t^{n-1}=
\zeta(2)\frac{\sum\_{n=1}^{\infty} \varphi(n)t^{n-1}}{\sum\_{n=1}^{\infty} nt^{n-1}}
$$
seems to be greater then 1 when $t$ increases to 1 (my caclulations say so), and analogous things appear for other averaging procedures involving $\varphi(n)$.


Does it have some sense and/or explanation?



Asymptotic propeties of Euler function
Write $S(t)= \sum \varphi(n)t^n$. A standard calculation gives


$S(e^{-u})= \frac{1}{2\pi i}\int\_{(3)}\frac{\zeta(s-1)}{\zeta(s)}\Gamma(s)u^{-s} ds$,


so pushing the contour to $Re(s)=3/2$ (say) gives $S(e^{-u})=\zeta(2)^{-1} u^{-2}+O(u^{-\frac{3}{2}})$ as $u\to 1$. How many $n$ are you using in your numerical calculations?


