Last time we ended with two new concepts.
Our key example was the Riemann zeta function, where we saw that $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_p \left(\frac{1}{1p^{s}}\right)\, .$$
We also saw that $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}\left(1\frac{1}{p^s}\right)=\prod_p (1p^{s})$$ wherever these make sense.
What was surprising about this is that the Euler products for the Riemann $\zeta$ function and this function are simply multiplicative inverses, e.g. $$\prod_p \frac{1}{1p^{s}}=1/\left(\prod_p 1p^{s}\right)\, .$$
0.607931911854226 0.607931911854226 
0.607927101854027 0.607927101854027 
Recall that $\zeta(2)=\frac{\pi^2}{6}$ (though before we just used this as a sum, and didn't call it $\zeta(2)$.
(Incidentally, though Euler calculated even values of $\zeta$, it was only in 1978 that $\zeta(3)$ was shown to be irrational, and then named after the man who proved this, Roger Apéry. To this day, it is only known that at least one of the next four odd values is irrational.)
0.607927101854027 0.607927101854027 
At least things agree to four digits when we approximate it, so this seems reasonable.
But last time, we said that
This is really a quite remarkable connection between the discrete/algebraic point of view and the analytic/calculus point of view.
Proof:
We can immediately apply this to calculate the Dirichlet series of $\phi$ in terms of the Riemann $\zeta$ function. We have three facts:
So the series for $\phi$ (call it $P$) and $u$ multiply to the series for $N$. But now we have three other facts.
That means $$\sum_{n=1}^\infty \frac{\phi(n)}{n^s}=\frac{\zeta(s1)}{\zeta(s)}\; .$$
1.36837198604112 1.36837198604112 
1.36843277762021 1.36843277762021 
It turns out that such Euler products (and hence nice computations like this) show up quite frequently.
Theorem: If $\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$ converges absolutely and $f$ is multiplicative, then $$\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots\right)\, .$$
The proof is essentially identical to what we proved with Moebius $\mu$'s Dirichlet series converging to its Euler product.
We can now announce four application facts which we can prove, using these tools. We will prove as many as we have time for, and leave the rest.
Let's prove these things.
Fact: $$\qquad \sum_{n=1}^{\infty}\frac{1}{p_n}$$ diverges, with $p_n$ the $n$th prime.
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Proof (divergence of "prime harmonic series"):
As with many other series, we will prove it by looking at the "tails" beyond a point that keeps getting further out. In this case, we'll choose the 'tail' beyond the first $k$ primes.
Now assume that in fact the "prime harmonic series" converges; we will derive a contradiction.
This contradiction occurred because we assume that the tail of the original series was finite, so the series in fact diverges!
Fact: The chances that a random integer lattice point is visible from the origin is $\frac{6}{\pi^2}$.
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Proof (probability of visibility):
First off, it should be pretty clear from the picture that if $(x,y)$ has a gcd, then $\left(\frac{x}{d},\frac{y}{d}\right)$ is right on the line of sight from the origin to $(x,y)$ so that it is blocked off  hence we are really just asking for the probability that two integers are relatively prime. So we will prove something about this instead.
Fact: The Dirichlet series for $\mu(n)$ is $\zeta(s)/\zeta(2s)$.
Proof:
It's nearly completely symbolic at this point!
Fact: $$\lim_{n\to\infty}\frac{\frac{1}{n}\sum_{k=1}^n \phi(k)}{\frac{3}{\pi^2}n}=1$$

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Proof (that the average of $\phi$ is linear with slope $3/\pi^2$):
We will crucially use these facts:
Now we will look at the summation function for $\phi$. We can once again think of it as summing things up in two different ways  along the hyperbola $xy=k$ and along each column.
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So $$\sum_{k=1}^n\sum_{d\mid k}\mu(d)\frac{k}{d}=\sum_{d=1}^n\mu(d)\left(\sum_{k=1}^{\lfloor \frac{n}{d}\rfloor}k\right)$$
So dividing by $n$ and taking the limit, we get the asymptotic $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\phi(n)=\lim_{n\to\infty}\frac{1}{2}\frac{n^2}{n}\frac{6}{\pi^2}=\lim_{n\to\infty}\frac{3}{\pi^2}n\; .$$
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