Is there a way to compute $\sum_{n=0}^\infty 1/(1+n!)$?

Question

Is there a way to compute the exact value of the following series? \begin{equation} \sum_{n=0}^\infty\frac{1}{1+n!} \end{equation}

I know that it converges to a number less than $e$, since $e=\sum_{n=0}^\infty 1/n!$. I also know that the approximated value is $1.52607$. But can I express the exact value using known constants or functions, such as $e, \pi, \Gamma$?

I know that $(1/n!) - (1/[1 + n!]) = 1/[n!(1+n!)],$ but that idea doesn't seem to lead anywhere. — user2661923, Sep 20 '20 at 03:57
What about integration. Again, this is just an idea. $\frac{D(1/n!)}{dn} = -\frac{n!}{1+n!}.$ — user2661923, Sep 20 '20 at 04:01
@Vilakshan Your comment begs a question which I (for one) have never heard of an answer to. How does one go about determining whether a specific (for example) infinite summation has a closed form expression? Related question: how does one prove (for example) that there is no closed form expression for $\int e^{(x^2)}dx$? — user2661923, Sep 20 '20 at 04:11
@user2661923 : For integrals, see the Risch algorithm and differential Galois theory (as used here). The situation for sums is more fragmented. A method for some sums is to show a sum is nonelementary by investigating its Wilf-Zeilberger pair. There are other methods (for other forms of sums) in the literature. — Eric Towers, Sep 20 '20 at 04:27

score 2 · Accepted Answer · answered Sep 20 '20 at 07:57

For a reasonable approximation $$\sum_{n=0}^\infty\frac{1}{1+n!}\sim\sum_{n=0}^p\frac{1}{1+n!}+\sum_{n=p+1}^\infty\frac{1}{n!}-\sum_{n=p+1}^\infty\frac{1}{(n!)^2}+\cdots$$ $$\sum_{n=p+1}^\infty\frac{1}{n!}=e\left(1-\frac{ \Gamma (p+1,1)}{\Gamma (p+1)}\right)$$ $$\sum_{n=p+1}^\infty\frac{1}{(n!)^2}=I_0(2)-a_p$$ where the $a_p$ make the sequence $$\left\{0,1,2,\frac{9}{4},\frac{41}{18},\frac{1313}{576},\frac{5471}{2400},\frac{118 1737}{518400},\frac{28952557}{12700800},\frac{1235309099}{541900800},\frac{15009 0055529}{65840947200}\right\}$$

Using $p=9$ $$\frac{10373124947763317933}{6797289565413518325}+e-\frac{98641}{36288}+\frac{150090055529}{65840947200}-I_0(2)$$ which gives $$1.5260681344733308247571$$ while the "exact" value is $$1.5260681344733308247780$$

Is there a way to compute $\sum_{n=0}^\infty 1/(1+n!)$?

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