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.

The key formula is $$\frac{\phi(n)}{n}=\sum_{d\mid n} \frac{\mu(d)}{d}$$
From this, you get that:
$$\begin{align}f(N)=&\sum_{k=1}^{N}\frac{\phi(k)}{k}\\&=\sum_{k=1}^{N}\sum_{d\mid k} \frac{\mu(d)}{d}\\
&= \sum_{d=1}^{N}\frac{\mu(d)}{d}\left\lfloor\frac N d\right\rfloor\end{align}$$
Then $\left|\left\lfloor\frac N d\right\rfloor - \frac{N}d\right|<1$ so $$\left|\frac{f(N)}N-\sum_{d=1}^{N}\frac{\mu(d)}{d^2}\right|<\frac{1}{N}\sum_{d=1}^{N}\frac1{d^2}< \frac{\zeta(2)}{N}$$
But we know that $$\sum_{d=1}^{\infty}\frac{\mu(d)}{d^2}=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$$
