Consequence of Bertrand's Postulate I read that the following can be proved using Bertrand's postulate (there's always a prime between $n$ and $2n$):
$\forall N\in\mathbb N$, there exists an even integer $k>0$ for which there are at least $N$ prime pairs $p$, $p+k$.
But I have no idea how to prove it. Any help would be much appreciated.
Many thanks.
 A: You don' need Bertrand's Postulate to prove it, it is enough with Euler's theorem on the divergence of $\sum \frac{1}{p}$. The argument is as follows:
For some integer $n$, consider the $n-2$ differences $p_{i+1}-p_i$ for $i=2,3,\dots, n-1$, if they take at most 
$$ T_n = \left\lfloor\frac{n-2}{N}\right\rfloor $$
different values then there is at least one taken $N$ times and we are done. 
Suppose otherwise that for every $n$ there are more than $T_n$ different values in the set $\{ p_{i+1}-p_i\,\,; i=2,3,\dots,n\}$ then
$$ p_n - p_2 = \sum_{i=2}^{n-1} p_{i+1}-p_i \ge 2+4+6+\dots+2T_n+2(T_n+1) = (T_n+1)(T_n+2) $$
but
$$ (T_n+1)(T_n+2) \ge \frac{(n-2)^2}{N^2}$$ 
So 
$$ \frac{1}{p_n} \le \frac{N^2}{(n-2)^2 + 3N^2} $$
if we fix $N$ and sum for $n=N,N+1,\dots,M $ we get
$$\sum_{i\le M} \frac{1}{p_i} \le -\sum_{i\le N} \frac{1}{p_i} + N^2 \sum_{N\le n \le N} \frac{1}{(n-2)^2 + 3N^2} $$
But the right hand side is bounded and by Euler's theorem the right hand isn't so for $n$ large enough we get a contradiction.
Note that this proves something stronger: for every $N$ there is an integer $k$ such that there are more than $N$ consecutive primes with difference $k$.
I read this argument in some old issue of the American Math Monthly (in 1985 or earlier), I'm sorry I can't give a reference.
A: It's not a direct consequence of the postulate. Indeed, consider the set $P = \{ 2^k : k \geq 0\}$. This set also satisfies Bertrand's postulate, but the differences $2^a - 2^b$ are unique.
