How to approximate $\sum_{k=1}^n k!$ using Stirling's formula?

Question

How to find summation of the first $n$ factorials,

$$1! + 2! + \cdots + n!$$

I know there's no direct formula, but how can it be estimated using Stirling's formula?

Another question :

Why can't we find the summation of n! ? Why there's no direct formula?

Answer 1

Stirling's formula gives us that $$n! \sim \sqrt{2 \pi n} \left( \dfrac{n}e\right)^n$$ i.e. $$\lim_{n \to \infty} \dfrac{n!}{\sqrt{2 \pi n} \left( \dfrac{n}e\right)^n} = 1$$ It is not hard to show that your sum, $$\sum_{k=1}^{n} k! \sim n!$$ and hence $$\sum_{k=1}^{n} k! \sim \sqrt{2 \pi n} \left( \dfrac{n}e\right)^n$$

EDIT To see that $\displaystyle \sum_{k=1}^{n} k! \sim n!$, note that \begin{align} \sum_{k=1}^{n} k! & = n! \left( 1 + \dfrac1n + \dfrac1{n(n-1)} + \dfrac1{n(n-1)(n-2)} + \cdots + \dfrac1{n!}\right)\\ & \leq n! \left( 1 + \dfrac1n + \dfrac{n-1}{n(n-1)}\right)\\ & = n! \left( 1 + \dfrac2n\right) \end{align} Hence, $\displaystyle \sum_{k=1}^{n} k! \sim n!$.

Answer 2

There is the direct formula:

$$\sum_{k=1}^{n-1} \Gamma(k)=(-1)^{n+1}\Gamma(n)(!(-n))+C$$

Where !(x) is subfactorial.