TITLE: Ratio of sum of Euler's totient to $n$: $\lim_{n \to \infty} {\log \left( \sum_{k=2}^n \varphi(k) \right) \over \log(n)}$
QUESTION [6 upvotes]: This is more a casual/recreational question... 
It seems to me, that the limit as given in the subject line
$$\lim_{n \to \infty}  {\log \left( \sum_{k=2}^n \varphi(k) \right) \over \log(n)} = \log_n \left( \sum_{k=2}^n \varphi(k) \right)  \le 2 $$
Possibly this is somehow trivial. And does it approach 2 in the limit?

REPLY [4 votes]: We have the asymptotic result $$\sum_{n\leq x}\phi(n)=\frac{3}{\pi^{2}}x^{2}+O\left(x\log x\right),$$ so your limit is $2$. 
For a complete overview with proof of the above as well as proofs of the $\Omega$ type results for the error term, take a look at this Blog Post.
A solution to this question can also be found in two other answers of mine on Math Stack Exchange:  Probability that two random numbers are coprime and Asymptotic formula for $\sum_{n\leq x}\mu(n)[x/n]^2$ and the Totient summatory function $\sum_{n\leq x} \phi(n)$