数学問題4番
問題
計算
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{3n^2+6n+2}{n^3+3n^2+2n}\biggr)^2\)
カッコ内
\(\displaystyle\frac{3n^2+6n+2}{n^3+3n^2+2n}\)
\(=\displaystyle\frac{n(n+1)+(n+1)(n+2)+n(n+2)}{n(n+1)(n+2)}\)
\(=\displaystyle\frac{1}{n}+\displaystyle\frac{1}{n+1}+\displaystyle\frac{1}{n+2}\)
全体
カッコ内の変形を代入して、まずは展開します。
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{n}+\displaystyle\frac{1}{n+1}+\displaystyle\frac{1}{n+2}\biggr)^2=\)
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{n^2}+\displaystyle\frac{1}{(n+1)^2}+\displaystyle\frac{1}{(n+2)^2}\)\(+\displaystyle\frac{2}{n(n+1)}+\displaystyle\frac{2}{n(n+2)}+\displaystyle\frac{2}{(n+1)(n+2)}\biggr)\)
最初の三項
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{n^2}+\displaystyle\frac{1}{(n+1)^2}+\displaystyle\frac{1}{(n+2)^2}\biggr)\) は
\(\displaystyle\sum_{n=1}^{\infty}\displaystyle\frac{1}{n^2}=\displaystyle\frac{\pi^2}{6}\) を使用する。
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{n^2}+\displaystyle\frac{1}{(n+1)^2}+\displaystyle\frac{1}{(n+2)^2}\biggr)\)
\(=\displaystyle\frac{\pi^2}{6}+\biggl(\displaystyle\frac{\pi^2}{6}-1\biggr)+\biggl(\displaystyle\frac{\pi^2}{6}-1-\displaystyle\frac{1}{4}\biggr)\)
\(=\displaystyle\frac{\pi^2}{2}-\displaystyle\frac{9}{4} \)
後半三項
部分分数分解を使います。
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{2}{n(n+1)}+\displaystyle\frac{2}{n(n+2)}+\displaystyle\frac{2}{(n+1)(n+2)}\biggr)\)
\(=2\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}\biggr)+\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+2}\biggr)+2\displaystyle\sum_{n=1}^{\infty}\biggl(\displaystyle\frac{1}{n+1}-\displaystyle\frac{1}{n+2}\biggr)\)
これらはすべて相殺しあい最初の方の数項のみ残ります。
\(=2+\biggl(1+\displaystyle\frac{1}{2}\biggr)+2\times\displaystyle\frac{1}{2}\) \(=\displaystyle\frac{9}{2}\)
答え
\(\displaystyle\sum_{n=1}^{\infty} \biggl(\displaystyle\frac{3n^2+6n+2}{n^3+3n^2+2n}\biggr)^2=\displaystyle\frac{\pi^2}{2}-\displaystyle\frac{9}{4}+\displaystyle\frac{9}{2}\) \(=\displaystyle\frac{\pi^2}{2}+\displaystyle\frac{9}{4}\)
