17. 令 \[S = \sum\limits_{k = 1}^n {\sqrt {k\left( {k + 2} \right)} } = \sum\limits_{k = 1}^n {\sqrt {{{\left( {k + 1} \right)}^2} - 1} } < \sum\limits_{k = 1}^n {\left( {k + 1} \right)} = \frac{{n\left( {n + 3} \right)}}{2}\]
又由 GM \(\geq\) HM
\[S = \sum\limits_{k = 1}^n {\sqrt {k\left( {k + 2} \right)} } \ge \sum\limits_{k = 1}^n {\frac{{k\left( {k + 2} \right)}}{{\left( {k + 1} \right)}}} = \sum\limits_{k = 1}^n {\frac{{{{\left( {k + 1} \right)}^2} - 1}}{{\left( {k + 1} \right)}}} = \sum\limits_{k = 1}^n {\left( {k + 1} \right)} - \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} \]
因此 \[\frac{3}{2} - \frac{{\sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} }}{n} < \frac{S}{n} - \frac{n}{2} < \frac{3}{2}\]
令 \[{H_n} = \sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} \] ，由Stolz-Cesaro Theorem，
\[\mathop {\lim }\limits_{n \to \infty } \frac{{{H_{n + 1}} - {H_n}}}{{\left( {n + 1} \right) - n}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + 2}} = 0 \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{H_n}}}{n} = 0\]
所以\[\mathop {\lim }\limits_{n \to \infty } \frac{3}{2} - \frac{{\sum\limits_{k = 1}^n {\frac{1}{{k + 1}}} }}{n} = \mathop {\lim }\limits_{n \to \infty } \frac{3}{2} = \frac{3}{2}\] ，由夾擠定理，
\[\mathop {\lim }\limits_{n \to \infty } \frac{S}{n} - \frac{n}{2} = \frac{3}{2}\]
[ 本帖最後由 ouchbgb 於 2023-4-14 13:21 編輯 ]
