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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 ?

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  • $\begingroup$ I mean by distributing the $1/n!$ we arrive at $$\sum_{j=1}^{20} \frac{j}{\prod_{k=0}^{j-1} (n-k)}$$ but I wouldn't say this is much prettier to work with $\endgroup$
    – WaveX
    Jan 27, 2020 at 5:25
  • $\begingroup$ I read $t$ as $t!$ earlier :P $\endgroup$
    – user721016
    Jan 27, 2020 at 5:59

1 Answer 1

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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 $n \geq 20$.

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