On proving the convergence of $1/n^2\sum_{1\le k\le n}\varphi(k)$ Let $$\Phi_n=\frac{1}{n^2}\sum_{k=1}^n\varphi(k).$$ How one can show that  $\Phi_n$ is convergent sequence? 
(Here, $\varphi$ denotes the  Euler's  totient function.)
And please, without any monster asymptotic estimations.
Thank you!
EDIT:
Can we actually prove the convergence without finding the limit?  

$\sum_{k=1}^n\varphi(k)$ counts pairs of integers $a,b$ with $a < b \le n$. So if we ignore the $a < b$ condition the thing we are taking the limit of is just the probability that if we take two numbers $a, b$ less than $n$ at random that they are relatively prime.  For large enough $n$ we have the probability that two numbers is divisible by a prime $p$ is very close to $1/p^2$.  Now doing an inclusion-exclusion over primes we get: $P(a,b \ \text{have a common factor}) = \sum_{p}p^{-2} - \sum_{p_1,p_2}p_1^{-2}p_2^{-2} + \sum_{p_1,p_2,p_3}p_1^{-2}p_2^{-2}p_3^{-2}-...$. So $P(a,b \text{ are relatively prime}) = (1-2^{-2})(1-3^{-2})... = 1/\zeta(2) = 6/ \pi^2$. Halving this to take into account the $a<b$ condition gives the desired result.  Yea, I'm being a bit fuzzy on the convergence details but this is definitely the idea.
