中級者
数学B：数列
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ヒント：１．部分分数分解を使う．２．(1)を使って，(2)を証明しよう．
問題
(1) 2以上の整数\(n\)に対し
\(\displaystyle \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n(n+1)}\)
を求めよ．
(2) 任意の正の整数\(n\)に対し
\(\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\)\(<2\)が成り立つことを示せ．(オリジナル）
解答
(1)\(\displaystyle \frac{1}{n(n+1)}\)\(=\displaystyle \frac{1}{n}-\frac{1}{n+1}\)
よって，\(\displaystyle \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{n(n+1)}\)\(=\displaystyle 1-\frac{1}{n+1}\)
(2)
\(n \geqq 2\)の時，\(n^2>n(n-1)\) \(\Leftrightarrow\) \(\displaystyle \frac{1}{n^2}<\frac{1}{n(n-1)}\)が成立する．
\(\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\)\(=\displaystyle \sum_{k=1}^{n} \frac{1}{k^2}\)\(=\displaystyle 1+\sum_{k=2}^{n} \frac{1}{k^2}\)\(<\displaystyle 1+\sum_{k=2}^{n} \frac{1}{k(k-1)}\)
(1)より，\(\displaystyle 1+\sum_{k=2}^{n} \frac{1}{k(k-1)}\)\(=\displaystyle 2-\frac{1}{n+1}\)\(<2\)
よって，\(\displaystyle \frac{1}{1^2}+\frac{1}{2^2}+\cdots+\frac{1}{n^2}\)\(<2\)が成り立つ．
参考入試問題
(1) 2以上の整数\(n\)に対し
\(\displaystyle \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{(n-1)n(n+1)}\)
を求めよ．
(2) 任意の正の整数\(n\)に対し
\(\displaystyle \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\cdots+\frac{1}{n^3}\)\(\displaystyle <\frac{5}{4}\)が成り立つことを示せ．(02 一橋大）
解答
(1)
\(\displaystyle \frac{1}{(n-1)n(n+1)}\)\(=\displaystyle \frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\)
よって，\(\displaystyle \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\cdots+\frac{1}{(n-1)n(n+1)}\)\(=\displaystyle \frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)+\frac{1}{2}\left(\frac{1}{2\cdot3}-\frac{1}{3\cdot4}\right)+\cdots+\frac{1}{2}\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\)\(=\displaystyle \frac{1}{4}-\frac{1}{2n(n+1)}\)
(2)
\(n \geqq 2\)の時，\(n^3>(n-1)n(n+1)\) \(\Leftrightarrow\) \(\displaystyle \frac{1}{n^3}<\frac{1}{(n-1)n(n+1)}\)が成立する．
\(\displaystyle \frac{5}{4}-\sum_{k=1}^{n}\frac{1}{k^3}\)\(=\displaystyle \frac{1}{4}-\sum_{k=2}^{n}\frac{1}{k^3}\)\(>\displaystyle \frac{1}{4}-\sum_{k=2}^{n}\frac{1}{(k-1)k(k+1)}\)\(=\displaystyle \frac{1}{2n(n+1)}>0\)
よって，\(\displaystyle \sum_{k=1}^{n}\frac{1}{k^3}\)\(<\displaystyle \frac{5}{4}\)