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设n为自然数，则 \[ n! = 1 \cdot 2 \cdot 3 \cdot \cdots \cdot n \] 称为n的阶乘．并且规定0!=1．
\[ \begin{array}{l} n! = \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n e^{\frac{\theta }{{12n}}} \left( {0 < \theta < 1} \right) \\ n! \approx \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n \left( 当n充分大 \right) \\ \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n < n! < \sqrt {2\pi n} \left( {\frac{n}{e}} \right)^n \left( {1 + \frac{1}{{12n - 1}}} \right) \\ \end{array} \]
\[ \begin{array}{l} \sum\limits_{j = 1}^n {j!j} = \left( {n + 1} \right)! - 1 \\ \sum\limits_{j = 1}^n {\frac{j}{{\left( {j + 1} \right)!}} = 1 - \frac{1}{{\left( {n + 1} \right)!}}} \\ \sum\limits_{j = 1}^n {\frac{{j^2 + j - 1}}{{\left( {j + 2} \right)!}} = \frac{1}{2} - \frac{{n + 1}}{{\left( {n + 2} \right)!}}} \\ \sum\limits_{j = 1}^n {\frac{{j2^j }}{{\left( {j + 2} \right)!}} = 1 - \frac{{2^{n + 1} }}{{\left( {n + 2} \right)!}}} \\ \end{array} \] \[ \begin{array}{l} \sum\limits_{j = 0}^n {\frac{1}{{j!\left( {n - j} \right)!}} = \frac{{2^n }}{{n!}}} \\ \sum\limits_{j = 0}^n {\left( { - 1} \right)^j \frac{{n\left( {n + j - 1} \right)!}}{{\left( {j!} \right)^2 \left( {n - j} \right)!}} = 0} \\ \sum\limits_{j = 1}^n {\frac{j}{{\left( {2j + 1} \right)!!}} = \frac{1}{2}\left[ {1 - \frac{1}{{\left( {2n + 1} \right)!!}}} \right]} \\ \sum\limits_{j = 1}^n {\frac{{\left( {2j - 1} \right)!!}}{{\left( {2j + 2} \right)!!}} = \frac{1}{2} - \frac{{\left( {2n + 1} \right)!!}}{{\left( {2n + 2} \right)!!}}} \\ \end{array} \]