極限の問題

こんにちは.今日は,面白いと感じた極限の問題を紹介しようと思います.最後まで読んでいってください.

では,解いていきましょう.極限の中身を$f(n)$とおきます.

${\displaystyle f(n)=\frac{(n+1{)}^{n+1}(n-1{)}^{n-1}-n^n\cdot n^n}{n^n(n-1{)}^{n-1}}}$

分母を少し変形すると

${\displaystyle \frac{1}{n^n(n-1{)}^{n-1}}=\frac{1}{n^n\cdot n^{n-1}}\cdot \frac{n^{n-1}}{(n-1{)}^{n-1}}}$

${\displaystyle =\frac{1}{n^{2n-1}}(1+\frac{1}{n-1}{)}^{n-1}}$

${\displaystyle \therefore f(n)=(1+\frac{1}{n-1}{)}^{n-1}\frac{(n+1{)}^{n+1}(n-1{)}^{n-1}-n^{2n}}{n^{2n-1}}}$

それでは,右側の分数の分子を$g(n)$として変形していきます.

${\displaystyle g(n)=(n+1{)}^2(n+1{)}^{n-1}(n-1{)}^{n-1}-n^{2n}}$

${\displaystyle =(n+1{)}^2(n^2-1{)}^{n-1}-n^{2n}}$

${\displaystyle =(n^2+2n+1)(n^2-1{)}^{n-1}-n^{2n}}$

${\displaystyle =(n^2-1+2n+2)(n^2-1{)}^{n-1}-n^{2n}}$

${\displaystyle =2n(1+\frac{1}{n})(n^2-1{)}^{n-1}+(n^2-1{)}^n-n^{2n}}$

${\displaystyle =2n(1+\frac{1}{n})(n^2-1{)}^{n-1}-\{(n^2{)}^n-(n^2-1{)}^n\}}$

${\displaystyle \therefore f(n)=(1+\frac{1}{n-1}{)}^{n-1}\frac{2n(1+\frac{1}{n})(n^2-1{)}^{n-1}-\{(n^2{)}^n-(n^2-1{)}^n\}}{n^{2n-1}}}$

ここで

${\displaystyle A_n=2n(1+\frac{1}{n})(n^2-1{)}^{n-1}\cdot \frac{1}{n^{2n-1}}}$

${\displaystyle B_n=\frac{(n^2{)}^n-(n^2-1{)}^n}{n^{2n-1}}}$

とすれば,求める極限は

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(1+\frac{1}{n-1}{)}^{n-1}\cdot (A_n-B_n)}$

になります.当然

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(1+\frac{1}{n-1}{)}^{n-1}=e}$

です.次に

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{A}_n=\underset{n\to \mathrm{\infty }}{lim}2n(1+\frac{1}{n})(n^2-1{)}^{n-1}\cdot \frac{1}{n^{2n-1}}}$

${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}2(1+\frac{1}{n})(n^2-1{)}^{n-1}\frac{1}{n^{2(n-1)}}}$

${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}2(1+\frac{1}{n})(1-\frac{1}{n^2}{)}^{n-1}}$

${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}\frac{2n^2}{n^2-1}(1+\frac{1}{n})(1-\frac{1}{n^2}{)}^n}$

${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}(2+\frac{2}{n^2-1})\cdot 1\cdot \{(1-\frac{1}{n^2}{)}^{n^2}{\}}^{\frac{1}{n}}}$

${\displaystyle =2\cdot 1\cdot 1=2}$

次に,$B_n$の極限を求めます.

${\displaystyle B_n=\frac{(n^2{)}^n-(n^2-1{)}^n}{n^{2n-1}}}$

${\displaystyle =\frac{1}{n^2}(n^2-n^2+1)\sum _{k=0}^{n-1}(n^2{)}^{n-1-k}(n^2-1{)}^k}$

${\displaystyle =\frac{1}{n^{2n-1}}{n}^{2(n-1)}\sum _{k=0}^{n-1}\frac{1}{n^{2k}}(n^2-1{)}^k}$

${\displaystyle =\frac{1}{n}\sum _{k=0}^{n-1}\frac{(n^2-1{)}^k}{n^{2k}}}$

${\displaystyle =\frac{1}{n}\sum _{k=0}^{n-1}(1-\frac{1}{n^2}{)}^k}$

${\displaystyle =\frac{1}{n}\sum _{k=0}^{n-1}{\phi }_n^k}$ (シグマの中身を${\phi }_n$としました)

${\displaystyle =\frac{1}{n}(1+{\phi }_n+{\phi }_n^2+\cdots +{\phi }_n^{n-1})}$

ここで,$1>{\phi }_n>{\phi }_n^2>\cdots >{\phi }_n^{n-1}>0$なので

${\displaystyle {\phi }_n^{n-1}<\frac{1}{n}(1+{\phi }_n+{\phi }_n^2+\cdots +{\phi }_n^{n-1})<1}$

${\displaystyle \therefore {\phi }_n^{n-1}<B_n<1}$

ここで

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{\phi }_n^{n-1}=\underset{n\to \mathrm{\infty }}{lim}(1-\frac{1}{n^2}{)}^{n-1}=1}$

ですから,挟み撃ちの原理より

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{B}_n=1}$

よって,求める極限は

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\{\frac{(n+1{)}^{n+1}}{n^n}-\frac{n^n}{(n-1{)}^{n-1}}\}=e(2-1)=e}$

となって答です.

最後まで読んでくれてありがとうございました.