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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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    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 $n^2$.2012-06-29
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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 http://oeis.org/A003422 which doesn't give any estimates but gives some formulas and links that might be helpful in getting estimations.2012-06-29
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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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    :Read your solution again,why is the summation equal to $n!$(i m sure,this is incorrect).2012-06-29
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    @avatar, Marvis says it's *asymptotic*, not equal, to $n!$.2012-06-29
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    @avatar Where did I say that the summation is $n!$?2012-06-29
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    @Marvis: Sorry, i ignored the asymptotic symbol2012-06-29
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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 $$ \sum\limits_{k = 1}^n {k!} \sim \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \frac{{13}}{{12n}} + \frac{{313}}{{288n^2 }} + \frac{{108041}}{{51840n^3 }} + \cdots } \right) . $$