Pi and the sum of reciprocals of primes?

Question

So I know that $$\sum_{\underset{\Large p\; prime}{p=1}}^{\infty}\frac{1}{p}$$ blows up. But doing some fun on mathematica I found out that when the sum isn't infinite, it was so close to $3$ and I conjectured that there exists an $\epsilon$ such that $$\sum_{\underset{\Large p\; prime}{p=1}}^{\epsilon}\frac{1}{p}=\pi$$ Is this conjecture true or false and more importantly why?

Answer 1

No, since if that sum were finite, i.e $n\in\mathbb N$ $$\dfrac{1}{p_1}+\dfrac{1}{p_2}+\dfrac{1}{p_3}+\cdots+\dfrac{1}{p_n},$$ then we can express it as: $a/b$ $(a,b\in\mathbb N)$ which is rational, unlike $\pi$.

Answer 2

It's false. $\pi$ is irrational, while the finite sum of rationals is rational.

Answer 3

No, the finite sum would be rational. Actually this idea gives an amusing proof that there must be infinitely many prime numbers. Suppose that there are only finitely many prime numbers. Then we have $\pi^2/6=\zeta(2)=\prod_{p}\frac{1}{1-1/p^2}$, which would be rational. Hence $\pi^2$ would be rational. This is a contradiction.