Does the sum of reciprocals of primes converge?
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 $\sum\limits_{p \leq x} \frac{1}{\sqrt{p}} \geq \frac{1}{2}\log{x} -\log{\log{x}}$
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!
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 = \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}
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.