Ta có:
\(\begin{array}{l}{\left( {1 + \dfrac{1}{n}} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{{.1}^{n - k}}.{{\left( {\dfrac{1}{n}} \right)}^k}} = \sum\limits_{k = 0}^n {C_n^k.\dfrac{1}{{{n^k}}}} \\ = 1 + C_n^1.\dfrac{1}{n} + C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}}\\ = 1 + 1 + C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}}\\ = 2 + C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}}\end{array}\)
Dễ thấy \(2 + C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}} > 2\) nên \({\left( {1 + \dfrac{1}{n}} \right)^n} > 2\)
Ta cần CM: \(C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}} < 1\)
Xét \(C_n^k.\dfrac{1}{{{n^k}}} = \dfrac{{n!}}{{k!\left( {n - k} \right)!.{n^k}}} = \dfrac{{n\left( {n - 1} \right)...\left( {n - k + 1} \right)}}{{{n^k}}}.\dfrac{1}{{k!}}\)
Mà \(n\left( {n - 1} \right)\left( {n - 2} \right)...\left( {n - k + 1} \right) < {n^k}\) nên \(\dfrac{{n\left( {n - 1} \right)...\left( {n - k + 1} \right)}}{{{n^k}}} < 1\)
\( \Rightarrow \dfrac{{n\left( {n - 1} \right)...\left( {n - k + 1} \right)}}{{{n^k}}}.\dfrac{1}{{k!}} < \dfrac{1}{{k!}} < \dfrac{1}{{k\left( {k - 1} \right)}} = \dfrac{1}{{k - 1}} - \dfrac{1}{k}\)
Do đó \(C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}} < 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{{n - 1}} - \dfrac{1}{n} = 1 - \dfrac{1}{n} < 1\)
Vậy \(2 + C_n^2.\dfrac{1}{{{n^2}}} + ... + C_n^n.\dfrac{1}{{{n^n}}} < 3\) hay \({\left( {1 + \dfrac{1}{n}} \right)^n} < 3\)
