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Let’s calculate the sum of factorials!

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Hello, guys! It’s me! Warotan!
So…Who are you?→ABOUT MEJ

Japanese version.

I write similar articles and will add contents, so please read more other articles.

Factorial of a natural number

A factorial of a natural number is defined as

\(\begin{eqnarray} n! \equiv n\left(n-1\right)\left(n-2\right)\cdots 3\cdot 2\cdot 1 \end{eqnarray}\)

Let us consider the summation of the factorials from 1 to \(n\),

\(\begin{eqnarray} S_{n} \equiv \sum_{k=1}^{n}k! \end{eqnarray}\)

Transforming \(k!\)

It is difficult to directly calculate the sum, so we do a little transforming.
With a permutation \(_{n}P_{k} = \frac{n!}{\left(n-k\right)!}\),

\(\begin{eqnarray} S_{n} &=& \sum_{k=1}^{n}k! \\ &=& \sum_{k=0}^{n-1}\left(n-k\right)! \\ &=& \sum_{k=0}^{n-1}n!/ _{n}P_{k} \\ &=& n!\sum_{k=0}^{n-1} 1/ _{n}P_{k} \end{eqnarray}\)

In fact, we can find this summation.

This is where the \(\Gamma\) function comes in

Do you remember the \(\Gamma\) function that I wrote about before? (English version coming soon)
Here, it is the turn of a variant, the incomplete \(\Gamma\) function.

The incomplete \(\Gamma\) function is defined as

\(\begin{eqnarray} \Gamma\left(a,x\right) \equiv \int_{x}^{\infty}{\rm e}^{-t}t^{a-1}{\rm d}t \end{eqnarray}\)

For \(a>0\), integrating by parts,

\(\begin{eqnarray} \Gamma\left(-a,x\right) &=& \int_{x}^{\infty}{\rm e}^{-t}\frac{1}{-a}\frac{{\rm d}}{{\rm d}t}t^{-a}{\rm d}t \\ &=& \frac{1}{-a}{\rm e}^{-t}t^{-a}|_{t\to\infty}-\frac{1}{a}{\rm e}^{-x}x^{-a}-\frac{1}{a}\int_{x}^{\infty}{\rm e}^{-t}t^{-a}{\rm d}t \\ &=& -\frac{1}{a}{\rm e}^{-x}x^{-a}-\frac{1}{a}\Gamma\left(-a+1,x\right) \end{eqnarray}\)

A recurrence formula appears.
Repeating this, we get

\(\begin{eqnarray} \Gamma\left(-a,x\right) &=& -\frac{1}{a}{\rm e}^{-x}x^{a}-\frac{1}{a\left(1-a\right)}{\rm e}^{-x}x^{1-a}-\frac{1}{a\left(1-a\right)\left(2-a\right)}{\rm e}^{-x}x^{2-a} \\ &-& \cdots -\frac{1}{a\left(1-a\right)\left(2-a\right)\cdots\left(n-a-1\right)}{\rm e}^{-x}x^{n-a-1} \\ &-& \frac{1}{a\left(1-a\right)\left(2-a\right)\cdots\left(n-a-1\right)}\Gamma\left(n-a,x\right) \\ &=& {\rm e}^{-x}\sum_{k=1}^{n}\left(-1\right)^{k}\frac{1}{ _{a}P_{k}}x^{k-a}-\Gamma\left(n-a,x\right) \end{eqnarray}\)

If \(a = n,\,x = 1\),

\(\begin{eqnarray} \Gamma\left(-n,-1\right) &=& {\rm e}\sum_{k=1}^{n}\left(-1\right)^{2k-n}\frac{1}{ _{n}P_{k}}-\Gamma\left(0,-1\right) \\ &=& \left(-1\right)^{n}{\rm e}\sum_{k=1}^{n}\frac{1}{ _{n}P_{k}}-\Gamma\left(0,-1\right) \end{eqnarray}\)

We did it!
It includes \( \sum_{k}\frac{1}{ _{n}P_{k}}\).
From this equation,

\(\begin{eqnarray} \sum_{k=1}^{n}\frac{1}{ _{n}P_{k}} = \left(-1\right)^{n}{\rm e}^{-1}\left[\Gamma\left(-n,-1\right)-\Gamma\left(0,-1\right)\right], \end{eqnarray}\)

and therefore,

\(\begin{eqnarray} S_{n} &=& n!\sum_{k=0}^{n-1}\frac{1}{ _{n}P_{k}} \\ &=& n!\left[\sum_{k=1}^{n}\frac{1}{ _{n}P_{k}}+\frac{1}{_{n}P_{0}}-\frac{1}{ _{n}P_{n}}\right] \\ &=& n!\left[\left(-1\right)^{n}{\rm e}^{-1}\left[\Gamma\left(-n,-1\right)-\Gamma\left(0,-1\right)\right]+1-\frac{1}{n!}\right] \\ &=& n!\left(-1\right)^{n}{\rm e}^{-1}\left[\Gamma\left(-n,-1\right)-\Gamma\left(0,-1\right)\right]+n!-1 \end{eqnarray}\)

Summary

Now, that’s all what I wanted to write.

This time, we found the summation of factorials.
It is the summation of integers, but the story didn’t fit within integers.

Thank you for reading and please spread this blog if you like.

If you have any comment, please let me know from the e-mail address below or the CONTACT on the menu bar.
tsunetthi(at)gmail.com
Please change (at) to @.

Or you can also contact me via twitter (@warotan3)

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