Ratio of sum of Euler's totient to $n$: $\lim_{n \to \infty} {\log \left( \sum_{k=2}^n \varphi(k) \right) \over \log(n)}$ 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?
 A: Thm: We have
$$
\sum_{n < x} \varphi(n) \sim \frac{3}{\pi^2} x^2
$$
Proof:
Since
$$
n = \sum_{d | n} \varphi(d)
$$
by Moebius inversion we get
$$
\varphi(n) = n \sum_{d | n} \frac{\mu(d)}{d}
$$
Therefore
$$
\sum_{n < x} \varphi(n) = \sum_{n < x} n \sum_{d | n} \frac{\mu(d)}{d}
$$
Interchanging summation we get
$$
\sum_{d < x} \frac{\mu(d)}{d} \sum_{d | n, n < x} n
$$
The inner sum equals
$$
d \cdot \frac{x^2}{2 d^2} + O(x) = \frac{x^2}{2d} + O(x)
$$
Therefore the final answer is
$$
\sum_{d < x} \frac{\mu(d)}{2 d^2} \cdot x^2 + O(x\log x) = 
\frac{1}{2\zeta(2)} x^2 + O(x\log x)
$$
because the later sum converges to $1 / \zeta(2) = 6/\pi^2$. 
$\square$
EDIT: In particular your limit is indeed equal to $2$!
EDIT 2: Actually for most integers $\varphi(n) \asymp n$. 
In fact the proportion of integers $n <  x$ such that $\alpha n < \varphi(n) < \beta n$, with $\alpha < 1$ converges to a continuous distribution function $$\mathbb{P}(\alpha < X < \beta) > 0$$ where explicitely $$ X := \prod_{p} \bigg ( 1 - \frac{X(p)}{p} \bigg )$$ and the $X(p)$ are independent random variables with $$\mathbb{P}(X(p) = 1) = \frac{1}{p} \text{ and } \mathbb{P}(X(p) = 0) = 1 - \frac{1}{p}.$$ 
This is Schoenberg's theorem.
A: 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 $\sum_{n\leq x} \phi(n)$
