What is the probability that the GCD of $n$ naturals $\le m$ is 1? What is the probability that the GCD of $n$ naturals $\le m$ is 1?
For example, if the there are $n=2$ numbers less than or equal to $m=2$, there are four combinations - {1,1},{1,2},{2,1},{2,2}.  Only the last combination, {2,2}, has a GCD greater than 1.  So with $n=2, m=2$, the probability is 3/4.
 A: (Note: it seems that I've used $k$ instead of $n$.)
This probability is given by function
$$C_k(m):=\frac1{m^k}\sum_{\substack{a_1,\dots,a_k\leq m\\\gcd(a_1,\dots,a_k)=1}}1$$
where the sum is extended to all the tuples $(a_1,\dots,a_k)$ with $a_i\leq m$ ($i=1,\dots,n$) that are relatively prime. One may prove that 
$$C_k(m)\sim\frac1{\zeta(k)};$$
that is to say, this probability tends to $1/\zeta(k)$ as $k\to\infty$ (where $\zeta(k)$ is the Riemann zeta function.) In particular, for $k=2$, the "probability" (more precisely, the asymptotic density) that two randomly chosen integers are relative prime is $6/\pi^2$. The proof of this result would be as follows:
$$\begin{align*}
C_2(m)&=\frac1{m^2}\sum_{a\leq m}\sum_{\substack{b\leq m\\(a,b)=1}}1
=\frac2{m^2}\left(\sum_{a\leq m}\sum_{\substack{b\leq a\\(a,b)=1}}1\right)-1
=\frac2{m^2}\sum_{a\leq m}\varphi(a)-1,
\end{align*}$$
where $\varphi(a)$ is the Euler's totient function, that counts the number of relatively prime to $a$ that are $\leq a$. In the second equality we made use of the symmetry: summing the pairs $(a,b)$ of the square $a,b\leq m$ such that $\gcd(a,b)=1$ is the same as summing the pairs inside the lower triangle $a\leq m,b\leq a$ and the pairs inside the upper triangle $b\leq m,a\leq b$ that satisfies $\gcd(a,b)=1$ (minus the intersection, which consists of a single point.) Both quantities turn out to be the same because $(a,b)=1$ iff $(b,a)=1$.
Now, since $\varphi=\mu*N$, we have (if we write $a=dq$)
$$\begin{align*}
\sum_{a\leq m}\sum_{d|a}\mu(a)\frac ad&=\sum_{d\leq m}\mu(d)\sum_{q\leq m/d}q\\
&=\sum_{d\leq m}\left(\frac{m^2\mu(d)}{2d^2}+O(m/d)\right)\\
&=\frac{m^2}{2}\sum_{d\geq1}\frac{\mu(d)}{d^2}+O\left(\sum_{d>m}\frac{m^2}{d^2}\right)+O(m\log(m))
\end{align*}$$
so $\sum_{a\leq k}\varphi(a)\sim Am^2/2$, where $A=\sum_{d\geq1}\mu(d)/d^2$. Therefore
$$\lim_{m\to\infty}C_2(m)=\lim_{m\to\infty}\frac2{m^2}\frac{m^2}{2}A=A.$$
Now we are left to find $A$. We note that
$$\sum_{d\geq1}\frac{\mu(d)}{d^2}\sum_{q\geq1}\frac1{q^2}
=\sum_{d,q\geq1}\frac{\mu(d)}{(dq)^2},$$
so if we write $n=dq$, we can rewrite it as
$$\sum_{n\geq1}\frac1{n^2}\sum_{dq=n}\mu(d)
=\sum_{n\geq1}\frac1{n^2}\sum_{d|n}\mu(d)
=\sum_{n\geq1}\frac1{n^2}I(n)$$
where $I(n)$ is the identity under Dirchlet convolution (i.e., $I(1)=1$ and $I(n)=0$ otherwise.) Therefore
$$A\sum_{n\geq1}\frac1{n^2}=1$$
so $A=1/\zeta(2)=6/\pi^2$. A similar (but longer) argument shows that $C_k$ converges to $1/\zeta(k)$. 
