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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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3  
For a very good estimate, we can estimate the sum of the last three terms. So $(n(n-1)+(n-1)+1)(n-2)!$, where we use Stirling for last term. The stuff in front is just $n^2$. –  André Nicolas Jun 29 '12 at 7:01
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Why should there be a direct formula? The number of summations you can write down far exceeds the number of direct formulas available, so most sums don't have an exact formula. Anyway, these numbers are tabulated at oeis.org/A003422 which doesn't give any estimates but gives some formulas and links that might be helpful in getting estimations. –  Gerry Myerson Jun 29 '12 at 7:13
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visit: mathworld.wolfram.com/FactorialSums.html –  Aang Jun 29 '12 at 7:14

3 Answers 3

up vote 5 down vote accepted

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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1  
:Read your solution again,why is the summation equal to $n!$(i m sure,this is incorrect). –  Aang Jun 29 '12 at 7:19
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@avatar, Marvis says it's asymptotic, not equal, to $n!$. –  Gerry Myerson Jun 29 '12 at 7:23
    
@avatar Where did I say that the summation is $n!$? –  user17762 Jun 29 '12 at 7:24
    
@Marvis: Sorry, i ignored the asymptotic symbol –  Aang Jun 29 '12 at 7:26
    
Stolz–Cesàro theorem may help to better understand this fact. –  Chris's sis Jun 29 '12 at 9:01

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 $$ \sum\limits_{k = 1}^n {k!} \sim \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \frac{{13}}{{12n}} + \frac{{313}}{{288n^2 }} + \frac{{108041}}{{51840n^3 }} + \cdots } \right) . $$

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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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