上級者
数学B：数列
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ヒント：数列よりもメインは極限です．
問題
(1) \(a_{1}=1\)，\(\displaystyle \frac{a_{1}+2a_{2}+\cdots+na_{n}}{n}=a_{n+1}\)で，数列\(a_{n}\)の一般項を求めよ．
(2) \(\displaystyle \lim_{n\rightarrow\infty}(a_{n})^{\frac{1}{n}}\)を求めよ．(オリジナル)
解説
(1) \(\displaystyle \frac{a_{1}+2a_{2}+\cdots+na_{n}}{n}=a_{n+1}\) \(\Leftrightarrow\) \(a_{1}+2a_{2}+\cdots+na_{n}=na_{n+1}\cdots\)①
\(a_{1}+2a_{2}+\cdots+(n+1)a_{n+1}=(n+1)a_{n+2}\cdots\)①'
①’-①から
\((n+1)a_{n+1}=(n+1)a_{n+2}-na_{n+1}\) \(\Leftrightarrow\) \(\displaystyle \frac{a_{n+2}}{a_{n+1}}=\frac{2n+1}{n+1}\cdots\)②
②を\(n\)を下げてかけていくと
\(\displaystyle \frac{a_{n}}{a_{n-1}}\times\frac{a_{n-1}}{a_{n-2}}\times\cdots\times\frac{a_{2}}{a_{1}}\)\(=\displaystyle \frac{2n-3}{n-1}\times\frac{2n-5}{n-2}\times\cdots\times\frac{3}{2}\times\frac{1}{1}\)
ここで，右辺の分子分母に\(2n-2,2n-4,\cdots,2\)をかけると
\(a_{n}=\displaystyle \frac{(2n-2)!}{(n-1)!}\times\frac{1}{(2n-2)(2n-4)\cdots 2}\)
また，\((2n-2)(2n-4)\cdots 2=2^{n-1}(n-1)!\)
∴\(a_{n}=\displaystyle \frac{(2n-2)!}{2^{n-1}\{(n-1)!\}^{2}}\)
(2) \(\displaystyle (a_{n})^{\frac{1}{n}}\)に\(\log\)をとると
\(\displaystyle \frac{1}{n}\log \frac{(2n-2)!}{2^{n-1}\{(n-1)!\}^{2}}\)
\(\log\displaystyle \frac{(2n-2)!}{2^{n-1}\{(n-1)!\}^{2}}\)について
\(\log\displaystyle \frac{(2n-2)!}{2^{n-1}\{(n-1)!\}^{2}}\)\(=\displaystyle \log\frac{n(n+1)\cdots(2n-2)}{1\times 2\times\cdots\times(n-1)}-(n-1)\log2\)\(=\displaystyle \log\frac{1\times(1+\frac{1}{n})\times\cdots\times(2-\frac{2}{n})}{\frac{1}{n}\times\frac{2}{n}\times\cdots\times(1-\frac{1}{n})}-(n-1)\log2\)\(=\displaystyle \sum_{k=0}^{n-2}\log(1+\frac{k}{n})-\sum_{k=1}^{n-1}\log\frac{k}{n}-(n-1)\log2\)
よって\(\displaystyle \lim_{n\rightarrow\infty}\log(a_{n})^{\frac{1}{n}}\)\(=\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=0}^{n-2}\log(1+\frac{k}{n})-\frac{1}{n}\sum_{k=1}^{n-1}\log\frac{k}{n}-\frac{n-1}{n}\log2\)\(=\displaystyle \int_{0}^{1}\log(1+x)dx-\int_{0}^{1}\log x dx-\log2\)
ここで，\(\displaystyle \int_{0}^{1}\log(1+x)dx=\cdots=2\log2 -1\)
\(\displaystyle \int_{0}^{1}\log x dx\)\(=\displaystyle \lim_{t\rightarrow 0}\int_{t}^{1}\log x dx\)\(=\displaystyle \lim_{t\rightarrow +0}(-1+t-t\log t)=-1\) （広義積分なので，高校範囲では解が与えられた状態で出る）
よって\(\displaystyle \lim_{n\rightarrow\infty}\log(a_{n})^{\frac{1}{n}}=\log2\)
以上より\(\displaystyle \lim_{n\rightarrow\infty}(a_{n})^{\frac{1}{n}}=2\)