Sum of reciprocals of primes factorial: $\sum_{p\;\text{prime}}\frac{1}{p!}$ 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?
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!
I have a formula which equals this, an infinite sum to be precise, but it is given in terms of a prime counting function.

Here it is:

If pi[n]= 'the number of primes up to and including the number n',

Then the sum of reciprocals of primes factorial equals:

∑pi[n]*n/(n+1)!,

where n goes from 2 to infinity.