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?

Solution 1:

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!

Solution 2:

Some quick notes on the approximation of this sum

This sum can be approximated in the obvious way (with Maple):

SrpfN := proc (N)
local s, p; s := 0;
for p from 2 to N do
if isprime(p) = true then
s := s+1/factorial(p)
end if end do;
s;
end proc;

The problem with the above is, however, the procedure "isprime" of Maple, which, in the best possible case is $O(n^k)$ for some constant $k$ (Primes is in P).

There is an alternate way to approximate it, using the PNT. Indeed, if we set $S_{rpf}(N)=\sum\limits_{n=2}^N\frac{1}{p_n!}$ then, because the sum is asymptotic, we can also set: $p_n\sim \frac{n}{\ln(n)}$ and then,

$$ \begin{align} S_{rpf}&=\sum_{n=2}^\infty\frac{1}{p_n!}\\ &=\lim\limits_{N\to\infty}\left(\sum\limits_{n=2}^N\frac{1}{p_n!}+\sum_{n=N+1}^\infty\frac{1}{\left(\frac{n}{\ln(n)}\right)!}\right)\\ &=\lim\limits_{N\to\infty}\left(S_{rpf}(N)+\sum_{n=N+1}^\infty\frac{1}{\Gamma\left(\frac{n}{\ln(n)}+1\right)}\right)\\ &=\lim\limits_{N\to\infty}\left(S_{rpf}(N)+\int_{x=N+1}^\infty\frac{dx}{\Gamma\left(\frac{x}{\ln(x)}+1\right)}\right)\\ \end{align} $$

Therefore, two useful approximants are given, as:

$$ \begin{align} S_{rpf1}(N,M)&=S_{rpf}(N)+\sum\limits_{n=N+1}^M\frac{1}{\Gamma\left(\frac{n}{\ln(n)}+1\right)}\text{ (1)}\\ S_{rpf2}(N,M)&=S_{rpf}(N)+\int\limits_{n=N+1}^M\frac{dx}{\Gamma\left(\frac{x}{\ln(x)}+1\right)}\text{ (2)}\\ \end{align} $$

Note that because the approximants are dependent on asymptotic behavior, it helps to move $N$ to some large number and leave the first part of the sum fixed, as: $S_{rpf}(N)$. This limits the use of the "isprime" calculator. We then use the following code:

N := 'N'; M := 'M'; n := 'n';#clear vars

px := proc (x) options operator, arrow;#PNT
x/ln(x)
end proc

Srpf1 := proc (N, M) options operator, arrow;
SrpfN(N)+Sum(1/GAMMA(px(n)+1), n = N+1 .. M)
end proc

Srpf2 := proc (N, M) options operator, arrow;
SrpfN(N)+Int(1/GAMMA(px(x)+1), x = N+1 .. M)
end proc

And now we verify that we have indeed fast convergence:

N := 100;
print('M', 'SrpfN(M)', 'Srpf1(N, M)', 'Srpf2(N, M)')
for M from N to N+10 do
print(M,evalf(SrpfN(M)),evalf(Srpf1(N, M)),evalf(Srpf2(N, M)));
od;

      M, SrpfN(M), Srpf1(N, M), Srpf2(N, M)
     100, 0.6751984379, 0.6751984379, 0.6751984379
     101, 0.6751984379, 0.6751984379, 0.6751984379
     102, 0.6751984379, 0.6751984379, 0.6751984379
     103, 0.6751984379, 0.6751984379, 0.6751984379
     104, 0.6751984379, 0.6751984379, 0.6751984379
     105, 0.6751984379, 0.6751984379, 0.6751984379
     106, 0.6751984379, 0.6751984379, 0.6751984379
     107, 0.6751984379, 0.6751984379, 0.6751984379
     108, 0.6751984379, 0.6751984379, 0.6751984379
     109, 0.6751984379, 0.6751984379, 0.6751984379
     110, 0.6751984379, 0.6751984379, 0.6751984379

Are approximations calculated this way faster? Indeed they are. Killing unneeded terms, getting rid of the constants and simplifying, with:

expr := factor(expand(simplify(
n*asympt(1/GAMMA(n/ln(n)+1), n, 1), trig, exp, ln)));

we get:

$$O\left(\sum\limits_{n=N+1}^M \Gamma\left(\frac{n}{\ln(n)}+1\right)^{-1}\right)\sim \frac{\sqrt{n\ln(n)}\cdot\ln(n)^{\left(\frac{n}{\ln(n)}\right)}}{e^{n-\frac{n}{\ln(n)}}}$$

Now setting $a_n$ equal to the above, we find $\lim\limits_{n\to\infty}\frac{a_n}{n}=0$, which shows immediately that the order of the sum and the integral in (1) and (2) is sublinear, hence sub-polynomial.

Therefore the above approximations are calculated faster than using only the "isprime" function, which is of polynomial time.