Good Estimate

$\rightarrow$ We may consider the sum $2 \sum_{n\leq x}\phi (n)-1$ , which $\phi(n)$ is the totient function. )

$\rightarrow$ When $m \leq n$ , there is $\sum_{n\leq x}\phi (n)$ number of pairs of $(m,n)$ .

$\rightarrow$ The case when $n \leq m$ is the same, since we have counted the pair $(1,1)$ twice, we minus one from the total, and we get the stated formula.

$\odot$ We now prove that $\sum_{n\leq x}\phi (n) = (3/ \pi ^2 ) x^2 + O( x logx )$ .

As $\sum_{d \mid n}\phi (d) = n$ , by Möbius inversion formula, $\phi(n) = \sum_{d \mid n}\mu (d) (n/d)$ .

To get a good estimation of $\sum_{n\leq x}\phi (n)$ , we proceed as following.

$\sum_{n\leq x}\phi (n)$

$= \sum_{n\leq x} \sum_{d \mid n} \mu (d) (n/d)$

Define $k=n/d$

$= \sum_{{d,k},and, {dk\leq x}} \mu (d) (k)$

$= \sum_{d\leq x} \mu (d) \sum_{k \leq x/d} (k)$

Since $\sum_{k \leq x/d} (k)$ is sum of integers less than or equal to $x$ ,

the sum is $( \left \lfloor x/d \right \rfloor ) ( \left \lfloor x/d \right \rfloor +1 ) /2$

$\rightarrow$ $\sum_{n\leq x}\phi (n)$ $= \sum_{d\leq x} \mu (d) ( \left \lfloor x/d \right \rfloor ) ( \left \lfloor x/d \right \rfloor +1 ) /2$

Since $\left \lfloor x/d \right \rfloor$ $= x/d + O(1)$ ,

So $( \left \lfloor x/d \right \rfloor ) ( \left \lfloor x/d \right \rfloor +1 )$ $= ( x/d + O(1) ) ( x/d + O(1) ) = x^2/d^2 + O(x/d)$

so $\sum_{d\leq x} \mu (d) ( \left \lfloor x/d \right \rfloor ) ( \left \lfloor x/d \right \rfloor +1 ) /2$

$= 1/2 (x^2) \sum_{d\leq x} \mu (d) /d^2 + O ( \sum_{d\leq x} (x/d) \mu (d) )$

$= 1/2 (x^2) \sum_{d\leq x} \mu (d) /d^2 + O ( \sum_{d\leq x} (x/d) \left | \mu (d) \right | )$

$= 1/2 (x^2) \sum_{d\leq x} \mu (d) /d^2 + O ( \sum_{d\leq x} (x/d) )$

$= 1/2 (x^2) \sum_{d\leq x} \mu (d) /d^2 + O ( x \sum_{d\leq x} (1/d) )$

$= 1/2 (x^2) \sum_{d\leq x} \mu (d) /d^2 + O ( x \int_{1}^{x} ( 1/d) {\mathrm{d} d}) )$

$= 1/2 (x^2) \sum_{d\leq x} \mu (d) /d^2 + O ( x log x)$

So as $x \rightarrow \infty$ ,

$\lim_{x\rightarrow \infty}( 2 \sum_{n\leq x}\phi (n)-1 ) / (x^2))$

$= \lim_{x\rightarrow \infty}[x^2 ( \sum_{d\leq x} \mu (d) /d^2 + O ( x log x)]/x^2$

$= \lim_{x\rightarrow \infty} ( \sum_{d\leq x} \mu (d) /d^2 ) + O ( x log x)]/x^2$

As $\sum_{d\leq x} \mu (d) /d^2 = 1/ \zeta(2) = 6/( \pi )^2$

so $\lim_{x\rightarrow \infty}( 2 \sum_{n\leq x}\phi (n)-1 / (x^2)) = 6/( \pi )^2$

where $\zeta (s)$ is the Riemann zeta function, and so the solution is $\mathfrak{6/( \pi )^2 }$ .

Let $A_p$ denote the event that at least one of the numbers is not divisible by a given prime $p$ . Then we can see that $\mathbb{P}[A_p] = 1 - \frac{1}{p^2}$ . The event that the two nonzero integers are relatively prime is just $\bigcap_p A_p$ , and the events are independent, so we have $\mathbb{P}\left[\bigcap_p A_p\right] = \prod_p \mathbb{P}[A_p] = \prod_p \left(1 - \frac{1}{p^2}\right) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2}.$