Sum of Reciprocals of Primes is Divergent

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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:

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

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


$(2): \quad \displaystyle \lim_{n \to \infty} \left({\ln \left({\ln n}\right) - C}\right) = + \infty$


Proof 1

By Sum of Reciprocals of Primes is Divergent: Lemma:

$\displaystyle \lim_{n \to \infty} \left({\ln \left({\ln n}\right) - \frac 1 2}\right) = +\infty$

$\Box$


It remains to be proved that:

$\displaystyle \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \ln \left({\ln n}\right) - \frac 1 2$

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

Let $n \ge 1$.

\(\displaystyle \prod_{p \mathop \le n} \left({1 - \frac 1 p}\right)^{-1}\) \(=\) \(\displaystyle \prod_{p \mathop \le n} \sum_{k \mathop = 0}^\infty \frac 1 {p^k}\) $\quad$ Sum of Infinite Geometric Progression $\quad$
\(\displaystyle \) \(\ge\) \(\displaystyle \sum_{k \mathop = 1}^n \frac 1 k\) $\quad$ Fundamental Theorem of Arithmetic $\quad$
\(\displaystyle \) \(>\) \(\displaystyle \int_1^n \frac 1 x \rd x\) $\quad$ Integral Test $\quad$
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \ln n\) $\quad$ $\quad$


Then:

\(\displaystyle \ln \prod_{p \mathop \le n} \left({1 - \frac 1 p}\right)^{-1}\) \(=\) \(\displaystyle \sum_{p \mathop \le n} \ln \left({1 - \frac 1 p}\right)^{-1}\) $\quad$ Sum of Logarithms $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{p \mathop \le n} \sum_{k \mathop = 1}^\infty \frac 1 {k p^k}\) $\quad$ Power Series Expansion for $\ln \left({1 + x}\right)$ $\quad$
\(\displaystyle \) \(<\) \(\displaystyle \sum_{p \mathop \le n} \frac 1 p + \sum_{p \le n} \frac 1 {2p^2} \left({\sum_{k \mathop = 0}^\infty \frac 1 {p^k} }\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{p \mathop \le n} \frac 1 {p \left({p - 1}\right)}\) $\quad$ Sum of Infinite Geometric Progression $\quad$
\(\displaystyle \) \(<\) \(\displaystyle \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{n \mathop = 2}^\infty \frac 1 {n \left({n - 1}\right)}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{p \mathop \le n} \frac 1 p + \frac 1 2\) $\quad$ Telescoping Series $\quad$
\(\displaystyle \sum_{p \mathop \le n} \frac 1 p\) \(>\) \(\displaystyle \ln \prod_{p \mathop \le n} \left({1 - \frac 1 p}\right)^{-1} - \frac 1 2\) $\quad$ $\quad$
\(\displaystyle \) \(>\) \(\displaystyle \ln \left({\ln n}\right) - \frac 1 2\) $\quad$ by $(1)$ $\quad$

$\blacksquare$


Proof 2

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 Progression, we have:

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

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:

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

It follows by taking reciprocals that:

$\displaystyle \prod_{k \mathop = 1}^n \left({1 - \frac 1 {p_k} }\right) < \frac 1 {\ln p_n}$

Taking logarithms of each side:

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


Next, note that the line $y = 2 x$ in the cartesian plane lies below the curve $y = \ln \left({1 + x}\right)$ on the interval $\left[{- \frac 1 2 \,.\,.\, 0}\right]$.

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

Thus it follows that:

$- \dfrac 2 {p_k} < \ln \left({1 - \dfrac 1 {p_k} }\right)$

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.

$\blacksquare$


Proof 3

Aiming for a contradiction, suppose the contrary.

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

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

Let:

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

and:

$\displaystyle S \left({r}\right) = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$

Let $S \left({r, j}\right)$ be the sum of all of the terms from $S \left({r}\right)$ 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:

$P \left({r}\right) = \left\{ {p_{k + 1}, p_{k + 2}, \ldots, p_{m \left({r}\right)} }\right\}$


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

$\displaystyle \left({\sum_{n \mathop = k \mathop + 1}^{m \left({r}\right)} \frac 1 {p_n} }\right)^j < \left({\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }\right)^j < \left({\frac 1 2}\right)^j$

and also by Sum of Infinite Geometric Progression:

$\displaystyle S \left({r}\right) = \sum_{j \mathop = 1}^r S \left({r, j}\right) < \sum_{j \mathop = 1}^r \left({\frac 1 2}\right)^j < 1$

for every $r$.


Finally notice that:

$\displaystyle S \left({r}\right) = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n\mathop = 1}^r \frac 1 n$

which implies that $S \left({r}\right)$ diverges towards $+\infty$ by Harmonic Series is Divergent, a contradiction.

$\blacksquare$


Proof 4

Aiming for a contradiction, suppose the contrary.

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

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

Let:

$\displaystyle M_x = \set {n \in \N: 1 \le n \le x \text { and } n \text { is not divisible by any primes greater than } p_k}$.

Notice that:

$\size {M_x} \le {2^k} \sqrt x$

because if:

$n = m^2r$

then

$m \le \sqrt x$

and there are at most $2^k$ distinct non-square prime compositions using primes less than $p_k$.

Now let:

$N_{i, x} = \set {n \in \N: 1 \le n \le x \text{ and } n \text { is divisible by } p_i}$

Notice:

$\displaystyle \size {\set {1, \dots, x} \setminus M_x} \le \sum_{i \mathop = k + 1}^\infty \size {N_{i, x} } < \sum_{i \mathop = k + 1}^\infty \dfrac x {p_i} \implies \dfrac x 2 < \size {M_x}$

Finally notice that whenever:

$x \ge 2^{2 k + 2}$

then it cannot be the case that both:

$\dfrac x 2 < \size {M_x} \le {2^k} \sqrt x$


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