a) Chứng minh bổ đề:
${P_n} - {P_{n - 1}} = \left( {n - 1} \right){P_{n - 1}}$
$\begin{array}{l} {P_n} - {P_{n - 1}}\\ = n! - \left( {n - 1} \right)!\\ = \left( {n - 1} \right)!n - \left( {n - 1} \right)!\\ = \left( {n - 1} \right)!\left( {n - 1} \right) = \left( {n - 1} \right){P_{n - 1}} \end{array}$
Áp dụng bổ đề trên ta được:
$\left\{ \begin{array}{l} {P_2} - {P_1} = \left( {2 - 1} \right){P_1}\\ {P_3} - {P_2} = \left( {3 - 1} \right){P_2}\\ {P_4} - {P_3} = \left( {4 - 3} \right){P_3}\\ ....\\ {P_n} - {P_{n - 1}} = \left( {n - 1} \right){P_{n - 1}} \end{array} \right.$
Cộng vế theo vế ta được:
$\begin{array}{l}
\Rightarrow {P_n} - {P_1} = {P_1} + 2{P_2} + 3{P_3} + ... + \left( {n - 1} \right){P_{n - 1}}\\
\Rightarrow {P_n} = 1 + {P_1} + 2{P_2} + 3{P_3} + ... + \left( {n - 1} \right){P_{n - 1}}
\end{array}$
b)
$\begin{array}{l} 1 + \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ... + \dfrac{1}{{n!}} < 3\\ \Leftrightarrow \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ... + \dfrac{1}{{n!}} < 2 \end{array}$
Ta có:
$\begin{array}{l} \dfrac{1}{{1!}} = 1\\ \dfrac{1}{{2!}} = \dfrac{1}{{1.2}} = 1 - \dfrac{1}{2}\\ \dfrac{1}{{3!}} = \dfrac{1}{{1.2.3}} = \dfrac{1}{2} - \dfrac{1}{3}\\ \dfrac{1}{{4!}} = \dfrac{1}{{1.2.3.4}} < \dfrac{1}{{3.4}} = \dfrac{1}{3} - \dfrac{1}{4}\\ ..............\\ \dfrac{1}{{n!}} = \dfrac{1}{{1.2.3...\left( {n - 1} \right)n}} < \dfrac{1}{{n - 1}} - \dfrac{1}{n}\\ \Rightarrow \dfrac{1}{{1!}} + \dfrac{1}{{2!}} + ... + \dfrac{1}{{n!}} < 1 + 1 - \dfrac{1}{n} < 2 \end{array}$
Vậy ta được điều phải chứng minh.
c)
$\begin{array}{l} \dfrac{1}{{\left( {n - 1} \right)!}} + \dfrac{1}{{\left( {n - 2} \right)!}}\\ = \dfrac{{\left( {n - 2} \right)! + \left( {n - 1} \right)!}}{{\left( {n - 1} \right)!\left( {n - 2} \right)!}} = \dfrac{{\left( {n - 2} \right)!\left[ {1 + \left( {n - 1} \right)} \right]}}{{\left( {n - 2} \right)!\left( {n - 1} \right)!}}\\ = \dfrac{n}{{\left( {n - 1} \right)!}} = \dfrac{{n.n}}{{\left( {n - 1} \right)!.n}} = \dfrac{{{n^2}}}{{n!}} \end{array}$
