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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?

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    visit: http://mathworld.wolfram.com/FactorialSums.html2012-06-29

3 Answers 3

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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, $\displaystyle \sum_{k=1}^{n} k! \sim n!$.

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    Stolz–Cesàro theorem may help to better understand this fact.2012-06-29
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There is the direct formula:

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

Where !(x) is subfactorial.

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To obtain better approximations, note that for large $n$, we have $ \sum\limits_{k = 1}^n {k!} = n!\left( {1 + \frac{1}{n} + \frac{1}{{n\left( {n - 1} \right)}} + \frac{1}{{n\left( {n - 1} \right)\left( {n - 2} \right)}} + \cdots } \right) = n!\left( {1 + \frac{1}{n} + \frac{1}{{n^2 }} + \frac{2}{{n^3 }} + \cdots } \right). $ Substituting Stirling's formula $ n! \sim \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \frac{1}{{12n}} + \frac{1}{{288n^2 }} - \frac{{139}}{{51840n^3 }} - \cdots } \right) $ yields $ \sum\limits_{k = 1}^n {k!} \sim \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \frac{{13}}{{12n}} + \frac{{313}}{{288n^2 }} + \frac{{108041}}{{51840n^3 }} + \cdots } \right) . $