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Japanese version.
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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.
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