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

Question

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*}

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!

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?

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.

A Manageable Series Expansion:

Converting the Round function to elementary functions, we have this form:

$$e\sum_{m=0}^\infty \sum_{n=2}^\infty \frac{\left(\frac e\pi\frac{\tan^{-1}\left(\tan\left(\frac\pi en!\right) \right)}{n!}\right)^m}{n!}$$

After a bit more work, we remove $!n$ from the denominator replacing it with gamma regularized $Q(a,z)$:

$$S=\sum_{m=0}^\infty\sum_{k=0}^n(-1)^k e^{k+1}\binom mk\sum_{n=2}^\infty\frac{(!n)^k}{n!^{k+1}}=\boxed{e\sum_{m,k=0}^\infty\binom mk(-1)^k\sum_{n=2}^\infty\frac{Q^k(n+1,-1)}{n!}}$$

Shown here.

Conclusion:

A closed form is optional, but good alternative representations for $S$ would also work. Please do not make up any new function. Alternatively, what is $\sum\limits_{n=2}^\infty\frac{Q^k(n+1,-1)}{n!}$?

Answer 1

I cannot see this one included above because it uses a much simpler function than those above, but it might be equivalent to one of them in the end. We start from the exponential generating function for subfactorial

$$E(x)=\frac{e^{-x}}{1-x}$$

We have

$$E(x)=\sum\limits_{n=0}^{\infty}!n\frac{x^n}{n!}$$

Now what is good about this is that differentiation gives us easily

$$E^{(n)}(0)=!n$$

or

$$\sum\limits_{n=2}^{\infty}\frac1{!n}=\sum\limits_{n=2}^{\infty}\frac1{E^{(n)}(0)}$$

There are many tools for dealing with differentials. As an example, we can use, from complicated, yet precise, the Lanczos derivatives formula

$$E^{(n)}(0)=\lim_{h\to 0}\frac{(2n+1)!}{2^{n+1}n!}\frac1{h^{n+1}}\int_{-h}^h E(x)P_n(\frac{x}{h})\mathrm dx$$

Lanczos’ generalized derivative for higher orders (10)

where $P_n(x)$ is a Legendre polynomial, which leads to a direct connection to the exponential integral, but equally Cauchy formula, to the simpler central difference formulas, etc. still noting that $E^{(n)}(0)$ is an integer, which gives a clear error margin necessary for each calculation.

Due to the reciprocal involved, none of the attempted resolutions leads to any true simplification of the expression. However, out of all the given options, this seems like the simplest one.

Notice that if we define the difference operator in this manner

$$\frac{1}{f^{(n)}(x)}=\frac{1}{\frac{d^n}{dx^n}f(x)}=\frac{1}{D^n}f$$

(alike $\frac1{D}\sin(x)=\frac1{\sin'(x)}=\frac1{\cos(x)}$)

we can write a very short symbolical expression

$$\sum\limits_{n=2}^{\infty}\frac1{!n}=\frac1{D(D-1)}E(0)$$

which is a shorthand symbolical notation for operators sum

$$(\frac1{D(D-1)})f=(\sum\limits_{n=2}^{\infty}\frac1{D^n})f$$