化為定積分

提供一種方法供參考

寫的匆忙有錯誤，麻煩版主刪了吧

[latex]\because \sum_{k=0}^{n}\frac{1}{\sqrt{(n-k+1)(k+1)}}> \frac{n}{\sqrt{(n+1)^2}}=1,(n \to \infty )[/latex]
[latex]\begin{align*}\sum_{k=0}^{n}\frac{1}{\sqrt{(n-k+1)(k+1)}}&=\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{2n}}+\frac{1}{\sqrt{3(n-1)}}+\cdots +\frac{1}{\sqrt{n+1}}\\\\&< \frac{n}{2}\cdot \frac{1}{\sqrt{(\frac{n}{2}+1)^2}}=1,(n \to \infty )\end{align*}[/latex]
[latex]\therefore \lim_{n \to \infty }\sum_{k=0}^{n}\frac{1}{\sqrt{(n-k+1)(k+1)}}=1.[/latex]，