Is the series $\sum_{1}^{\infty}\frac{1}{p_{j}}$ where $p_{j}$ is the $j$th prime convergent? Does the series $\sum_{n=1 }^{\infty}1/p_{j} $ of  reciprocal primes converge?
Experimentally, it seems convergent.
 A: Here is an easy proof of the divergence of the series of prime reciprocals, which I saw in the American Mathematical Monthly in a bygone millennium. Assume for a contradiction that the series converges. Choose $N$ so that
$$\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\lt1.$$
Then
$$S=1+\left(\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\right)+\left(\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\right)^2+\cdots$$
is a convergent geometric series.
Let $d=P_1P_2\cdots P_N$ and consider the series
$$\frac11+\frac1{1+d}+\frac1{1+2d}+\frac1{1+3d}+\cdots$$
which of course diverges. However, each denominator $1+nd$, since it is not divisible by any of the first $N$ primes (note the resemblance to Euclid's proof), is a product of primes $\ge P_{N+1}$. Thus each term $\frac1{1+nd}$ occurs in the geometric series $S$ when the powers are expanded, that is,
$$\frac11+\frac1{1+d}+\frac1{1+2d}+\cdots\lt1+\left(\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\right)+\left(\frac1{P_{N+1}}+\cdots\right)^2+\cdots\lt\infty$$
and this contradiction proves the theorem.
If anyone reading this knows the source of this proof, please edit the attribution into my answer. Thanks.
The proof is given by Clarkson [Clarkson, James A. ( 1966) On the series of prime reciprocals. Proc. Amer. Math. Soc .. /7: 541; MR 32. #5573.]
https://www.ams.org/journals/proc/1966-017-02/S0002-9939-1966-0188132-7/S0002-9939-1966-0188132-7.pdf
A: In this answer, it is shown that
$$
\sum_{\substack{p\le n\\p\text{ prime}}}\frac1p=\log(\log(n))+O(1)
$$
and therefore, the sum diverges.
