Sum of Reciprocals of Primes is Divergent/Proof 2

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Theorem

The series:

$\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$

where:

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

is divergent.


Proof

Let $n \in \N$ be a natural number.

Let $p_n$ denote the $n$th prime number.

Consider the product:

$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$


By Sum of Geometric Sequence, we have:

\(\displaystyle \frac 1 {1 - \frac 1 2}\) \(=\) \(\displaystyle 1 + \frac 1 2 + \frac 1 {2^2} + \cdots\)
\(\displaystyle \frac 1 {1 - \frac 1 3}\) \(=\) \(\displaystyle 1 + \frac 1 3 + \frac 1 {3^2} + \cdots\)
\(\displaystyle \frac 1 {1 - \frac 1 5}\) \(=\) \(\displaystyle 1 + \frac 1 5 + \frac 1 {5^2} + \cdots\)
\(\displaystyle \) \(\cdots\) \(\displaystyle \)
\(\displaystyle \frac 1 {1 - \frac 1 {p_n} }\) \(=\) \(\displaystyle 1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots\)

Consider what happens when all these series are multiplied together.

A new series will be generated whose terms consist of all possible products of one term selected from each of the series on the right hand side.

This new series will converge in any order to the product of the terms on the left hand side.

By the Fundamental Theorem of Arithmetic, every integer greater than $1$ is uniquely expressible as a product of powers of different primes.

Hence the product of these series is the series of reciprocals of all (strictly) positive integers whose prime factors are no greater than $p_n$.

In particular, all (strictly) positive integers up to $p_n$ have this property.

So:

$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
\(\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}\) \(\ge\) \(\displaystyle \sum_{k \mathop = 1}^{p_n} \frac 1 k\)
\(\displaystyle \) \(>\) \(\displaystyle \int_1^{p_n + 1} \dfrac {\mathrm d x} x\)
\(\displaystyle \) \(=\) \(\displaystyle \map \ln {p_n + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \ln p_n\)

It follows by taking reciprocals that:

$\displaystyle \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$

Taking logarithms of each side:

$(1): \quad \displaystyle \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$


Next, note that the line $y = 2 x$ in the cartesian plane lies below the curve $y = \map \ln {1 + x}$ on the interval $\closedint {-\frac 1 2} 0$.

Also note that all primes are greater than or equal to $2$.

Thus it follows that:

$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$

Applying this to $(1)$ yields:

$\displaystyle -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$

and so:

$\displaystyle \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$

But:

$\displaystyle \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$

and so the series:

$\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$

is divergent.

$\blacksquare$


Historical Note

This proof of the divergence of the reciprocals of the primes was discovered by Leonhard Paul Euler.


Sources