What is the summation of some pseudo factorial?

Question

Given $n$ is a positive integer where $n>2$, what is the summation of the following?

$$S_n=1+n+n(n-1)+n(n-1)(n-2)+\cdot\cdot\cdot+n!$$

Answer 1

If you use Don Thousand's comment $$S_n=1+n\, S_{n-1} \qquad \text{with} \qquad S_0=1$$ or Ng Chung Tak's answser, you have $$S_n=e\, \Gamma (n+1,1)$$ where appears the incomplete gamma function.

This corresponds to sequence $A000522$ in $OEIS$ and it has, as generating function $\frac{e^x}{1-x}$.

The great Ramanujan gave as asymptotics $$S_n=e\,n!-\frac 1{n}+\frac 1{n^3}-\frac 1{n^4}-\frac 1{2n^5}+\frac 1{9n^6}+\cdots$$

Answer 2

Using the exact expression of remainder term for Taylor series expansion:

$$f(x)=\sum_{k=1}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^{k}+ \int_{a}^{x} \frac{(x-t)^{n}}{n!} f^{(n+1)}(t) \, dt$$

\begin{align} S_n &= n! \left[ \frac{1}{n!}+\frac{1}{(n-1)!}+\ldots+\frac{1}{2!}+\frac{1}{1!} \right] \\ &= n! \left[ e-1-\frac{1}{n!} \int_{0}^{1} (1-t)^{n} e^{t} \, dt \right] \end{align}