Sum of Reciprocals of Primes is Divergent/Proof 3

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Theorem

Let $n \in \N: n \ge 1$.


There exists a (strictly) positive real number $C \in \R_{>0}$ such that:

$(1): \quad \ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - C$

where $\Bbb P$ is the set of all prime numbers.


$(2): \quad \ds \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = +\infty$


Proof

Aiming for a contradiction, suppose the contrary.

If the prime reciprocal series converges then there must exist some $k \in \N$ such that:

$\ds \sum_{n \mathop = k \mathop + 1}^{\infty} \frac 1 {p_n} < \frac 1 2$

Let:

$\ds Q = \prod_{i \mathop = 1}^k {p_i}$

and:

$\ds \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$

Let $\map S {r, j}$ be the sum of all of the terms from $\map S r$ for which $1 + i Q$ has exactly $j$ prime factors.

Notice that $1 + i Q$ is coprime with every prime factor in $Q$.

Thus every prime factor of $1 + i Q$ where $i = 1, \ldots, r$ falls into some finite sequence of consecutive primes:

$\map P r = \set {p_{k + 1}, p_{k + 2}, \ldots, p_{\map m r} }$




Notice again that each term of $\map S {r, j}$ occurs at least once in the expansion of:

$\ds \paren {\sum_{n \mathop = k \mathop + 1}^{\map m r} \frac 1 {p_n} }^j < \paren {\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }^j < \paren {\frac 1 2}^j$

and also by Sum of Infinite Geometric Sequence:

$\ds \map S r = \sum_{j \mathop = 1}^r \map S {r, j} < \sum_{j \mathop = 1}^r \paren {\frac 1 2}^j < 1$

for every $r$.


Finally notice that:

$\ds \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n \mathop = 1}^r \frac 1 n$

which implies that $\map S r$ diverges towards $+\infty$ by Harmonic Series is Divergent, a contradiction.

$\blacksquare$


Historical Note

This proof is credited to James Andrew Clarkson.