Conjecture: $\lim_{N \to +\infty}\frac{1}{N}\sum_{k=1}^{N}\frac{\phi(k)}{k}=\frac{6}{\pi^2}$ I was playing around with the series $f(N)=\frac{1}{N}\sum_{k=1}^{N}\frac{\phi(k)}{k}$ and I found with Wolfram that $f(10,000)=0.607938$, which I noticed was very close to $\frac{6}{\pi^2}$.  
I am led to make the following

Conjecture: $\lim_{N \to +\infty}\frac{1}{N}\sum_{k=1}^{N}\frac{\phi(k)}{k}=\frac{6}{\pi^2}$

Well, is it true?
Note that its obvious that the sum is bounded above by $1$ (since $\phi(k)/k<1$), so it definitely doesn't diverge to infinity. Its also almost always decreasing. So it most likely converges.

Here's a start: 
It is well known that (a,b) = 1 for random a,b with probability $6/\pi^2.$ Then $\phi(k) \sim k \cdot \frac{6}{\pi^2}$ for large k, and because the density of primes is zero, by the definition of $\phi(x),$ so $$\frac{1}{n}\sum_{n\geq k\geq 1}\frac{\phi(k)}{k}\sim \frac{1}{n}\sum_{n\geq k\geq 1} \frac{6}{\pi^2}=\frac{6}{\pi^2}.$$ However this is not very rigorous.
