Finding $\lim_{n\to\infty}\Phi(n)/n^2$, When $\Phi(n)=\sum_{i=1}^{n}\phi(n)$ This exercise is meant to be 'explored' computationally. However, I implemented it in C++ and did not get anything better than a sequence of pseudo-random numbers.

Let $\Phi(n)=\sum_{i=1}^{n}\phi(n)$. Investigate the value of $\Phi(n)/n^2$ for increasingly large values of $n$, such as $n=100$, $n=1000$, and $n=10000$. Can you make a conjecture about the limit of this ratio as $n$ grows large without bound?

Notice that $\Phi(n)=n\phi(n)$. Hence, $\Phi(n)/n^2=\phi(n)/n$. Moreover, the largest value $\phi(n)/n$ ever attains is $1$ at $n=1$; everything else falls within the interval $(0,1)$, and the closest it gets to $1$ again is when $n$ is prime (since $\phi(p)=p-1$, and $(p-1)/p\approx1$ for very large primes $p$).
However, I am tempted to say that this function diverges, and that no conjecture about its limit can be concluded as a result.
What do you guys think?
 A: There is a very old result that says
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n \varphi(k)}{n^2}=\frac{3}{\pi^2}.$$
The error term I have in notes is $O(x(\log x)^{2/3}(\log\log x)^{4/3})$, but undoubtedly there have been improvements on that.  There is a large literature. 
Added: The OP quoted correctly the textbook source of the problem, which asks about the behaviour of $(\sum_{i=1}^n\varphi(n))/n^2$.  This is undoubtedly a typo, since $\sum_{i=1}^n\varphi(n)=n\varphi(n)$. 
The ratio $\dfrac{\varphi(n)}{n}$ certainly bounces around a lot, and can be made arbitrarily close to $0$, and, much more easily, arbitrarily close to $1$.
A: Here is a detailed note regarding the Totient Summatory function.  Part 1 and 2 should be of interest, and in part 2 there is a short proof.
Also see this Math Stack Exchange question and answer.
A: i found an interesting connection between the following probability problem and your question:
Question: If 2 integers are randomly chosen between 1 and n, then as n tends to infinity, find the probability that the chosen numbers are coprime.
Now the probability that 2 numbers are not divisible by a prime p is $(1-1/p^2)$. So we require the infinite product:
$$ \prod_{p}{(1-\frac{1}{p^2})}$$
And this can be easily seen to be $\frac{1}{\zeta(2)}= \frac{6}{\pi^2}$ which is the answer, where $ \zeta(z) $is the Reimann zeta function.
Thinking in terms of definition, probability is ratio of favourable outcomes and total outcomes.
Now, number of possible pairs of numbers is $ n²$.
What is the number of coprime pairs $(x,y)$,$ 1≤x≤y≤n$, x,y being integers?
Setting $y=k$, the possible pairs is $\phi(k)$, because $ \phi(k) $is the number of integers less than k and prime to k.
So, required answer is just:
$$ \lim_{n\to \infty}  \frac{2(\phi(1)+\phi(2)+...\phi(n))}{n^2}$$
(2 introduced so that  if $(x,y) $are coprime, then and$ (y,x)$ is also coprime.)
So the twice the limit in your question is the probability that 2 chosen numbers in 1 to n is coprime!
So your answer must be
$$ \lim_{n\to \infty}  \frac{(\phi(1)+\phi(2)+...\phi(n))}{n^2} =\frac{1}{2\zeta(2)}= \frac{3}{\pi^2}$$
