一道求极限题

求极限

$$\lim\limits_{n\rightarrow \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{k(n-k+1)}}.$$

解：由

$$\sum_{k=1}^{n}\frac{1}{\sqrt{\frac{k}{n}(1-\frac{k-1}{n})}}\leq \sum_{k=1}^{n}\int_{\frac{k-1}{n}}^{\frac{k}{n}}\frac{dx}{\sqrt{x(1-x)}}=\int_{0}^{1}\frac{dx}{\sqrt{x(1-x)}}=\pi$$ $$\sum_{k=1}^{n}\frac{1}{\sqrt{\frac{k}{n}(1-\frac{k-1}{n})}}=\sum_{k=1}^{n}\frac{1}{\sqrt{\frac{k}{n}(1+\frac{2}{n}-\frac{k+1}{n})}}$$ $$\geq \sum_{k=1}^{n}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\frac{dx}{\sqrt{x(1+\frac{2}{n}-x)}}=\int_{\frac{1}{n}}^{1+\frac{1}{n}}\frac{dx}{\sqrt{x(1+\frac{2}{n}-x)}}$$ $$=\int_{\frac{1}{n}}^{1+\frac{2}{n}}\frac{dx}{\sqrt{x(1+\frac{2}{n}-x)}}-\int_{0}^{1+\frac{1}{n}}\frac{dx}{\sqrt{x(1+\frac{2}{n}-x)}}-\int_{1+\frac{1}{n}}^{1+\frac{2}{n}}\frac{dx}{\sqrt{x(1+\frac{2}{n}-x)}}$$ $$=\int_{0}^{1}\frac{dt}{\sqrt{t(1-t)}}-2\int_{0}^{\frac{1}{n}}\frac{dx}{\sqrt{x(1+\frac{2}{n}-x)}}(t=\frac{x}{1+\frac{2}{n}})$$ $$\geq \pi-\frac{2}{\sqrt{1+\frac{1}{n}}}\int_{0}^{\frac{1}{n}}\frac{1}{\sqrt{x}}dx=\pi-\frac{2}{\sqrt{1+\frac{1}{n}}}\frac{2}{\sqrt{n}}=\pi-\frac{4}{\sqrt{n+1}}$$

于是

$$\pi-\frac{4}{\sqrt{n+1}}\leq \sum_{k=1}^{n}\frac{1} {\sqrt{k(n-k+1)}}=\sum_{k=1}^{n}\frac{1}{\sqrt{\frac{k}{n}(1-\frac{k-1}{n})}}\leq \pi$$

从而极限是$\pi$.