Evaluating $\sum\limits_{x=2}^\infty \frac{1}{!x}$ in exact form.

$$\large \text{Introduction:}$$

We know that:

$$\sum_{x=0}^\infty \frac{1}{x!}=e$$

But what if we replaced $$x!$$ with $$!x$$ also called the subfactorial function also called the $$x$$th derangement number? This interestingly is just a multiple of $$e$$ and an Incomplete Gamma function based sum. This will get us a new number as the $$n=0$$ and 1 terms diverge as a result of the reciprocal. I also use the Generalized Exponential Integral function and the Round function. The OEIS entry for the constant is A281682:

\begin{align*} S &=\sum_{x=2}^\infty \frac{1}{!x} =\sum_{n=2}^\infty \frac{1}{\operatorname{Round}\bigl(\frac{x!}{e}\bigr)} =e\sum_{x=2}^\infty\frac{1}{Γ(x+1,-1)}= \\ &= e\sum_{n=3}^\infty \frac 1{Γ(x,-1)}=\sum_{X=0}^\infty \sum_{x=2}^\infty\frac{1}{Γ(X+1)Γ(x+1,-1)} =-e\sum_{x=2}^\infty\frac{(-1)^x}{E_{-x}(-1)}= -e\sum_{x=-\infty}^{-2}\frac{(-1)^x}{E_x(-1)} =1.63822707… \end{align*}

$$\large \text{Possible Abel-Plana formula Application:}$$

We can also use the Abel-Plana formula, and the alternate series version, to find an integral representation of the sum. You can also use other representations of the summand, but this integral is probably hard to work with.

Note the Abel-Plana formula may not work with the constant:

\begin{align*} S &=\sum_{x=0}^\infty\frac{1}{!(x+2)} =\frac{1}{2} + \int_0^\infty \frac{dx}{!(x+2)} + i\int_0^\infty\frac{\frac{1}{!(2+ix)}-\frac{1}{!(2-ix)}}{e^{2\pi x}-1} \, dx \\ \implies -\frac{S}{e} &=\sum_{x=0}^\infty\frac{(-1)^x}{E_{-x-2}(-1)}=-\frac{1}{2e} + \frac i2\int_0^\infty \left[\frac{1}{E_{-ix-2}(-1)}-\frac{1}{E_{ix-2}(-1)}\right] \operatorname{csch}(\pi x) \, dx \end{align*}

See this nice closed form result of

$$\sum_{x=-\infty}^0 \text {Im}(!x)=-\frac{\pi}{e^2}$$

I do not think this simple looking problem has been posted so far.

The sum does not need to be in closed form.

You also can rewrite it in terms of a better sum. I am more looking for an evaluation or manipulation of the sum. Please correct any mistakes and give me feedback!

$$\large \text{A Mittag-Leffler Insight:}$$

Because $$\sum_{x=2}^\infty \frac{1}{!x} =e\sum_{x= 2}^\infty\frac{1}{Γ(x+1,-1)}= e\sum_{n=3}^\infty\frac 1{Γ(x,-1)}$$

one may notice the relation to the Mittag-Leffler function:

$$\text E_{a,b}(x)=\sum_{n=0}^\infty \frac{x^n}{Γ(ax+b)}$$

The only problem is if there existed a function for the incomplete gamma function analogue of the Mittag-Leffler function. Maybe one can find this function or use the already known one?

$$\large \text{Another Integral Representation:}$$

It can be shown that the following is true using @Jack Barber’s method in

$$\sum_\Bbb N \text{erfc}(x)$$

Here is an integral representation using the linearity of the Floor function and the Meijer G function:

$$\sum_2^\infty \frac{1}{!x}=-\int_2^\infty \lfloor x-1\rfloor \frac{d}{dx} \frac{1}{!x}dx=\int_2^\infty \frac{d}{dx} \frac{1}{!x} dx-\int_2^\infty\lfloor x\rfloor \frac{d}{dx} \frac{1}{!x} dx=\frac1{!\infty}-\frac1{!2}-\int_2^\infty \lfloor x \rfloor \frac{d}{dx} \frac{1}{!x} =-1-\int_2^\infty \lfloor x \rfloor \frac{d}{dx} \frac{1}{!x}dx= -1-\int_2^\infty \lfloor x \rfloor\left(-\frac{\text G_{2,3}^{3,0}\big(-1\left|_{0,0,x+1}^{\ \ \ \ 1,1}\right)}{e(!x)^2}-\frac{i\pi}{!x}\right)dx =\frac1e\int_2^\infty \frac{\lfloor x\rfloor\text G_{2,3}^{3,0}\left(-1\big|_{0,0,x+1}^{\ \ \ \ 1,1}\right)}{(!x)^2} dx+i\pi\int_2^\infty\frac{\lfloor x\rfloor}{!x}dx-1$$

The Meijer G function is hard to use, but you can come up with many more integral representations using alternate forms of the floor function; there is even one in terms of elementary functions.

$$\large \text{Conclusion:}$$

A closed form is optional, but good alternative representations for S would also work. Please do not make up any new function.

• $\sum\frac1{!x}$ converges at the same rate as $\sum\frac1{x!}$, which is already faster than exponential; asking for even better convergence seems very ambitious! Aug 6 at 20:30
• @AlBrown Please see this definition from Wolfram Functions. From -1 to infinity diverges. Aug 6 at 21:21
• @AlBrown Does this help? I would have done a wolfram alpha link, but it refuses to work, so here is a results photo. Notice the end result is the same. Aug 6 at 21:29
• Plus i saw what you added. I meant integral from -1, but im good Aug 6 at 22:12
• And what does "a better form" mean? I know there's not much at OEIS; I take that to mean there's not much known. Aug 22 at 12:53