Sum of Reciprocals of Primes is Divergent/Proof 1

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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

By Sum of Reciprocals of Primes is Divergent: Lemma:

$\ds \lim_{n \mathop \to \infty} \paren {\map \ln {\map \ln n} - \frac 1 2} = +\infty$

$\Box$


It remains to be proved that:

$\ds \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - \frac 1 2$

Assume all sums and product over $p$ are over the set of prime numbers.

Let $n \ge 1$.

\(\ds \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1}\) \(=\) \(\ds \prod_{p \mathop \le n} \sum_{k \mathop = 0}^\infty \frac 1 {p^k}\) Sum of Infinite Geometric Sequence
\(\ds \) \(\ge\) \(\ds \sum_{k \mathop = 1}^n \frac 1 k\) Fundamental Theorem of Arithmetic
\(\ds \) \(>\) \(\ds \int_1^n \frac 1 x \rd x\) Cauchy Integral Test
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \ln n\)


Then:

\(\ds \ln \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1}\) \(=\) \(\ds \sum_{p \mathop \le n} \map \ln {1 - \frac 1 p}^{-1}\) Sum of Logarithms
\(\ds \) \(=\) \(\ds \sum_{p \mathop \le n} \sum_{k \mathop = 1}^\infty \frac 1 {k p^k}\) Power Series Expansion for $\map \ln {1 + x}$
\(\ds \) \(<\) \(\ds \sum_{p \mathop \le n} \frac 1 p + \sum_{p \mathop \le n} \frac 1 {2 p^2} \paren {\sum_{k \mathop = 0}^\infty \frac 1 {p^k} }\)
\(\ds \) \(=\) \(\ds \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{p \mathop \le n} \frac 1 {p \paren {p - 1} }\) Sum of Infinite Geometric Sequence
\(\ds \) \(<\) \(\ds \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{n \mathop = 2}^\infty \frac 1 {n \paren {n - 1} }\)
\(\ds \) \(=\) \(\ds \sum_{p \mathop \le n} \frac 1 p + \frac 1 2\) Definition of Telescoping Series
\(\ds \sum_{p \mathop \le n} \frac 1 p\) \(>\) \(\ds \ln \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1} - \frac 1 2\)
\(\ds \) \(>\) \(\ds \map \ln {\ln n} - \frac 1 2\) by $(1)$

$\blacksquare$



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