TITLE: Is there a way to compute $\sum_{n=0}^\infty 1/(1+n!)$?
QUESTION [6 upvotes]: 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 $e, \pi, \Gamma$?

REPLY [2 votes]: 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
$$1.5260681344733308247780$$