Does the sum of reciprocals of primes converge? Is this series known to converge, and if so, what does it converge to (if known)?
Where $p_n$ is prime number $n$, and $p_1 = 2$,
$$\sum\limits_{n=1}^\infty \frac{1}{p_n}$$
 A: Let's start with three lemmas:

*

*Suppose $A\subseteq\{1,2,3,\ldots\}$ and $\sum\limits_{n\in A} \dfrac 1 n < \infty$. Then $\sum\limits_{n\in B} \dfrac 1 n <\infty$ where $B$ is the closure of $A$ under multiplication.


*The closure of the set of primes under multiplication is all of $\{1,2,3,\ldots\}$.


*$\sum\limits_{n=1}^\infty \dfrac 1 n = \infty$.
The second lemma is obvious.  The third has a number of well known simple proofs. Here is one of those:
\begin{align}
& \frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \cdots \tag 1 \\[10pt]
= {} &\left(\frac 1 1 + \frac 1 2\right) + \left(\frac 1 3 + \frac 1 4\right) + \left(\frac 1 5 + \frac 1 6\right) + \cdots \\[10pt]
\ge {} & \left(\frac 1 2 + \frac 1 2 \right) + \left( \frac 1 4 + \frac 1 4 \right) + \left( \frac 1 6 + \frac 1 6 \right) + \cdots \tag 2 \\[10pt]
= {} & \frac 1 1 + \frac 1 2 + \frac 1 3 + \cdots
\end{align}
The inequality on line $(2)$ is strict if the sum on line $(1)$ is finite, and that leads us to a contradiction. ${}\qquad\blacksquare$
The proof of lemma 1 is most of the work; here it is:
\begin{align}
& \sum_{n\in B} \frac 1 n \le \overbrace{\sum_{\begin{smallmatrix} C\subseteq A \\[2pt] C \text{ is finite} \end{smallmatrix}} \prod_{k\in C} \frac 1 k = \prod_{a\in A} \sum_{x=0}^\infty \frac 1 {a^x}}^\text{factoring -- see below} = \prod_{a\in A} \frac 1 {1-\frac 1 a} \\[10pt]
= {} & \exp \sum_{a\in A} - \log\left( 1 - \frac 1 a\right) \le \exp \sum_{a\in A} \frac 1 a < \infty.
\end{align}
(As "Pipicito" points out in a comment below, some members of the set $B$ may occur more than once in the sum below and that is why $\text{“}{\le}\text{''}$ rather than $\text{“}{=}\text{''}$ should appear in the first step above.)
Here's the factorization in more detail: Let $A=\{a_1,a_2,a_3,\ldots\}$.  Then the product to the right of $\text{“}{=}\text{''}$ under the $\overbrace{\text{overbrace}}$ above is
\begin{align}
& \left( 1 + \frac 1 {a_1} + \frac 1 {a_1^2} + \frac 1 {a_1^3} + \cdots \right) \\
\times {} & \left( 1 + \frac 1 {a_2} + \frac 1 {a_2^2} + \frac 1 {a_2^3} + \cdots \right) \\
\times {} & \left( 1 + \frac 1 {a_3} + \frac 1 {a_3^2} + \frac 1 {a_3^3} + \cdots \right) \\
\times {} & \quad \cdots \cdots \\
\vdots~
\end{align}
When you expand the product, you multiply a term from the first factor, a term from the second factor, a term from the third factor, etc., but all except finitely many of those are $1$. The reason all but finitely many are $1$ is that if you multiply infinitely many non-$1$s, then the product is $0$, since its a product of infinitely many positive numbers less than $1/2$.  Then you add up all possible such finite products, and that gives you the sum to the left of $\text{“}=\text{''}$ under the $\overbrace{\text{overbrace}}$ above.
A: No, it does not converge. See this: Proof of divergence of sum of reciprocals of primes.
In fact it is known that $$\sum_{p \le x} \frac{1}{p} = \log \log x + A + \mathcal{O}(\frac{1}{\log^2 x})$$
Related: Proving $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$
A: I would like to note that this implies that according the Müntz-Szász Theorem that every continuous function in $[0,1]$ is a uniform limit of polynomials whose exponents are prime numbers!
