The series
$$\sum_{p\;\text{prime}}\frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\cdots$$
diverges as is well known. How about the following?
$$\sum_{p\;\text{prime}}\frac{1}{p!}=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+\frac{1}{11!}+\cdots$$
Its convergence is easily obtained by comparing with the series $e=\sum_{n=0}^\infty\frac{1}{n!}$. Numerically, we get
$$\sum_{p\;\text{prime}}\frac{1}{p!}\simeq 0.675198$$
Can we find the specific value?
Prime Numbers – Sum of Reciprocals of Primes Factorial
prime numberssequences-and-series
Best Answer
I calculated your number to 15 decimal places and entered it into the Inverse Symbolic Calculator. The only match was for the sum $\sum_{p}\frac{1}{p!}$.
Therefore, I would venture to guess that this number is unlikely to admit a simple description in terms of any known constants.
Interesting question, disappointing answer. Sorry!