How to evaluate the closed form of the sum with factorials?

Question

I have got the sum

$$\sum_{t = 1}^{20} t \cdot \frac{(n - t)!}{n!}$$

Is it possible to get the value of this in terms of $n$ without calculator ?

Answer 1

Yes it's possibile.

But you need a very good knowledge of special functions for this, in particular: Euler Gamma Function (complete, and incomplete).

Your sum is known in its closed for as $t$ goes to infinity:

$$\sum_{t = 0}^{+\infty} t\cdot \frac{(n-t)!}{n!} = \frac{1}{n!} \sum_{t = 0}^{+\infty} t\cdot (n-t)! = \frac{(-1)^n \Gamma (1-n)}{e}+\frac{(-1)^{n+2} n \Gamma (-n,-1)}{e}-1$$

Where $e$ is the Euler number; $\Gamma(-n)$ is the Euler Gamma Function and $\Gamma(a, b)$ is the incomplete Gamma function.

For your series, $t$ starts from one so we easily get the same result as before, since for $t = 0$ the very first term of the sum is zero.

$$\frac{1}{n!} \sum_{t = 1}^{+\infty} t\cdot (n-t)! = \frac{(-1)^n \Gamma (1-n)}{e}+\frac{(-1)^{n+2} n \Gamma (-n,-1)}{e}-1$$

From here is rather possible by hands to calculate the terms (tedious but doable):

$$\frac{1}{n!} \sum_{t = 1}^{20} t\cdot (n-t)! = \frac{20 (n-20)!}{n!}+\frac{19 (n-19)!}{n!}+\frac{18 (n-18)!}{n!}+\frac{17 (n-17)!}{n!}+\frac{16 (n-16)!}{n!}+\frac{15 (n-15)!}{n!}+\frac{14 (n-14)!}{n!}+\frac{13 (n-13)!}{n!}+\frac{12 (n-12)!}{n!}+\frac{11 (n-11)!}{n!}+\frac{10 (n-10)!}{n!}+\frac{9 (n-9)!}{n!}+\frac{8 (n-8)!}{n!}+\frac{7 (n-7)!}{n!}+\frac{6 (n-6)!}{n!}+\frac{5 (n-5)!}{n!}+\frac{4 (n-4)!}{n!}+\frac{3 (n-3)!}{n!}+\frac{2 (n-2)!}{n!}+\frac{(n-1)!}{n!}$$

Now just plug a value for $n$ and have fun with the calculation!

Beware that you need $n\in \mathbb{N}$ and here $n \geq 20$.