CMR với mọi số nguyên dương n>=2, ta có 2<(1+1/n)^n <3 20/08/2021 Bởi Kaylee CMR với mọi số nguyên dương n>=2, ta có 2<(1+1/n)^n <3
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 \({\left( {1 + \dfrac{1}{n}} \right)^n} < 3\) Bình luận
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 \({\left( {1 + \dfrac{1}{n}} \right)^n} < 3\)