[数Ⅲ]区分求積法の計算(その２)
問題〔(4)電気通信大2020年〕[目標10分]
(1) \(\displaystyle\lim\limits_{n\to\infty}\lbrace\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}＋･･･+\frac{n}{(n+n)^2}\rbrace\)
(画面サイズによって式がスワイプできます↑)
(2) \(\displaystyle\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}＋･･･+\frac{1}{2n-1})\)
(画面サイズによって式がスワイプできます↑)
(3) \(\displaystyle\lim\limits_{n\to\infty}\sum\limits_{k=1}^{2n}\log(\frac{k}{n})^\frac{1}{n}\)
(4) \(\displaystyle\lim\limits_{n\to\infty}\sum\limits_{k=n+1}^{2n}\frac{n}{k^2+3kn+2n^2}\)
【解答】
(1) \(\displaystyle\lim\limits_{n\to\infty}\lbrace\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}＋･･･+\frac{n}{(n+n)^2}\rbrace\)
\(\displaystyle＝\lim\limits_{n\to\infty}\frac{1}{n}\lbrace\frac{n^2}{(n+1)^2}+\frac{n^2}{(n+2)^2}＋･･･\frac{n^2}{(n+n)^2}\rbrace\)
\(\displaystyle＝\lim\limits_{n\to\infty}\frac{1}{n}\lbrace(\frac{1}{1+\frac{1}{n}})^2+(\frac{1}{1+\frac{2}{n}})^2＋･･･+(\frac{1}{1+\frac{n}{n}})^2\rbrace\)
\(\displaystyle＝\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^n\frac{1}{(1+\frac{k}{n})^2}\)
\(\displaystyle=\int_{0}^{1}\frac{1}{(1+x)^2}dx\)
\(\displaystyle=\int_{0}^{1}(1+x)^{-2}dx\)
\(\displaystyle=\left[-(1+x)^{-1}\right]_{0}^{1}\)
\(\displaystyle=\frac{1}{2}\)
(2)\(\displaystyle\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}＋･･･+\frac{1}{2n-1})\)
\(\displaystyle＝\lim\limits_{n\to\infty}(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}＋･･･+\frac{1}{n+(n-1)})\)
\(\displaystyle2n-1\)を\(\displaystyle n+(n-1)\)と見られるかがポイントだよ
\(\displaystyle＝\lim\limits_{n\to\infty}\frac{1}{n}\lbrace1+\frac{1}{1+\frac{1}{n}}+\frac{1}{1+\frac{2}{n}}＋･･･+\frac{1}{1+\frac{n-1}{n}}\rbrace\)
\(\displaystyle\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}\frac{1}{(1+\frac{k}{n})}\)
\(\displaystyle=\int_{0}^{1}\frac{1}{1+x}dx\)
\(\displaystyle=\left[\log|1+x|\right]_{0}^{1}\)
\(\displaystyle=\log2\)
(3) \(\displaystyle\lim\limits_{n\to\infty}\sum\limits_{k=1}^{2n}\log(\frac{k}{n})^\frac{1}{n}\)
\(\displaystyle=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{2n}\frac{1}{n}\log(\frac{k}{n})\)
\(\displaystyle=\int_{0}^{2}\log{x}dx\)
\(\displaystyle=\left[x\log{x}-x\right]_{0}^{2}\)
\(\displaystyle=2\log{2}-2\)
(4) \(\displaystyle\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=n+1}^{2n}\frac{n}{k^2+3kn+2n^2}\)
\(\displaystyle＝\lim\limits_{n\to\infty}\sum\limits_{k=n+1}^{2n}(\frac{1}{k+n}-\frac{1}{k+2n})\)
\(\displaystyle＝\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=n+1}^{2n}(\frac{1}{\frac{k}{n}+1}-\frac{1}{\frac{k}{n}+2})\)
\(\displaystyle=\int_{1}^{2}(\frac{1}{x+1}-\frac{1}{x+2})dx\)
\(\displaystyle=\left[\log|x+1|-\log|x+2|\right]_{1}^{2}\)
\(\displaystyle=2\log3-\log2\)
まとめ
区分求積法の計算は，シグマの上端と下端が常に\(\displaystyle k=1からn\)とは限りません．
しっかり演習しておきましょう！
