Deduce average order of $\phi(n)/n$ from probability that two integers are coprime

Question

I've seen proofs of the fact that the probability of two random integers being coprime is $\frac{6}{\pi^2}$ (all of them leading to a use of the Riemann Zeta function and the Basel problem). In several cases it was mentioned that one can easily deduce from that the fact that the average order of $\phi(n)/n$ is also equal to $\frac{6}{\pi^2}$. To me this is not clear, maybe I am missing something obvious. Is this deduction possible and if it is, is there a easy way to see it?

Answer 1

So the rigorous statement that "two random integers have a $6/\pi^2$ probability of being coprime" is $$ \lim_{N\to\infty}\frac{1}{N^2}\sum_{n=1}^N\sum_{m=1}^N \mathbf 1_{m\text{ and }n\text{ are coprime}}\to \frac 6{\pi^2}. $$

This is the same as $$ \lim_{N\to\infty}\frac 2{N^2}\sum_{n=1}^N \sum_{m=1}^{n-1}\mathbf 1_{m\text{ and }n\text{ are coprime}}\to \frac 6{\pi^2}. $$

Or

$$ \lim_{N\to\infty}\frac 2{N^2}\sum_{n=1}^N \phi(n)\to \frac 6{\pi^2}. $$ Using summation by parts, we have: $$ \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N\frac{\phi(n)}n= \lim_{N\to\infty} \frac 1N \left( \frac {S_N}N + \sum_{n=1}^{N-1}\frac{S_n}{n(n+1)}\right), $$ where $S_n=\phi(1)+\ldots+\phi(n)$. The previous equality shows $S_N/N^2\to 3/\pi^2$, so that the limit on the right of the displayed equation converges to $6/\pi^2$ as required.

Answer 2

As $\phi(n)$ counts the number of integers co-prime to it from $1$ to $n-1$, we would expect there to be $\dfrac{6}{\pi^2}$ of them.