Looking for a Simple Proof of the Divergence of the Prime Harmonic Series 
It is well known that $$\sum_{i=1}^{\infty} \frac{1}{p_{i}}$$ diverges, where the $p_{i}$'s are the prime numbers.

Does anybody know a very elementary proof of this result that would be suitable for calculus-level students?
Many thanks. 
 A: There is a very nice proof (due to Erdös)  that involves a minimum amount of calculus, it is the following:
If $\sum 1/p$ converges then there is an $n$ such that
$$1/p_{n+1}+1/p_{n+2} + \dots < \frac{1}{2}$$
chose one $n$ with this property and consider the set $A$ of all integers composed only by the first $n$ primes: $p_1,p_2,\dots,p_n$.
let $A(x)$ be the number of elements of $A$ smaller than $x$, now you can write eny element  $a \in A$ as $a=m^2s$ with $s$ squarefree and there are only $2^n$ squarefree integers made by the first $n$ primes. So for all $x$
$$ A(x) \le 2^n\sqrt{x} $$
On the other hand $x-A(x)$ counts the number of integer up to $x$ with at least one prime factor $\ge p_{n+1}$, but this number is at most
$$ \left \lfloor\frac{x}{p_{n+1}} \right \rfloor +
\left \lfloor\frac{x}{p_{n+2}} \right \rfloor + \dots \le \frac{x}{p_{n+1}} +\frac{x}{p_{n+2}} + \dots < \frac{x}2$$
so we get the two inequalities:
$$ A(x) > \frac x2 \quad \text{and} \quad A(x) \le 2^n\sqrt{x} $$
and combining both we get for all $x$:
$$ 2^{n+1} > \sqrt{x} $$
but this is clearly false if $x \ge 2^{2n+2}$ leading to a contradiction, so $\sum 1/p$ diverges.
A: Define $\zeta(s) = 1+ \frac{1}{2^s}+ \frac{1}{3^s}+ \frac{1}{4^s}+\cdots ,$ where $s \in \mathbb{R}, s> 1$.
We know that $\zeta(s) = \prod\limits_{p \text{ prime}} \left( 1 - \dfrac{1}{p^s} \right)^{-1}.$
Taking $\log$ of both the sides, we have:
$\log \zeta(s) = -\sum\limits_{p \text{ prime}}\log \left( 1 - \dfrac{1}{p^s} \right).$
Expanding $\log$ of right-hand side, we have:
$\log \zeta(s) = \sum\limits_{p \text{ prime}} \left( \dfrac{1}{p^s} \right)+ A(s).$
The term $A(s)$ takes into account the higher powers of primes. Moreover $A(s)$ converges at $s=1$
Taking $\ \displaystyle \lim_{s\to 1}$ , the left hand side is infinity and $A(s)$ is finite, so $\sum\limits_{p \text{ prime}} \left( \dfrac{1}{p} \right)$ diverges.
A: Note:
I improved my answer
to show that
$\sum_{p \le n}\frac1{p}
\ge \ln(\ln(n))-1
$.
Or Euler's
(off the top of my head):
By unique factorization
$\begin{array}\\
\sum_{k=1}^n \dfrac1{k}
&\le \dfrac1{\prod_{p \le n}(1-1/p)}\\
&=\prod_{p \le n}(1+\frac1{p-1})\\
&\le\prod_{p \le n}e^{\frac1{p-1}}
\qquad\text{since } e^x \ge 1+x\\
&=e^{\sum_{p \le n}\frac1{p-1}}\\
\text{so}\\
\sum_{p \le n}\frac1{p-1}
&\ge \ln(\sum_{k=1}^n \dfrac1{k})\\
&\ge \ln(\ln(n))
\qquad\text{by integral test}\\
\end{array}
$
Since
$\dfrac1{p-1}
\le \dfrac{2}{p}$,
we have
$\sum_{p \le n}\frac1{p}
\gt \frac12\ln(\ln(x))
$.
Here is an improvement
on this last paragraph.
$\begin{array}\\
\sum_{p \le n}\frac1{p-1}
-\sum_{p \le n}\frac1{p}
&=\sum_{p \le n}(\frac1{p-1}-\frac1{p})\\
&=\sum_{p \le n}\frac1{p(p-1)}\\
&\le\sum_{k=2}^{n}\frac1{k(k-1)}\\
&\lt 1\\
\end{array}
$
so
$\sum_{p \le n}\frac1{p}
\gt \sum_{p \le n}\frac1{p-1}-1
\ge \ln(\ln(n))-1
$.
