已知 $n\in\mathbb N^{\ast}$，求证：$\displaystyle\dfrac{2n}3\sqrt n<\sum_{k=1}^n\sqrt k<\dfrac{4n+3}6\sqrt n$．
解析 考虑裂项\[a_n=\left(n+\lambda\right)\sqrt n,n\in\mathbb N^{\ast},\]计算\[\begin{split} \dfrac{\sqrt n}{a_n-a_{n-1}}&=\dfrac{\sqrt n}{\left(n+\lambda\right)\sqrt n-\left(n-1+\lambda\right)\cdot \sqrt{n-1}}\\ &=\dfrac{\sqrt n}{\left(\sqrt n-\sqrt{n-1}\right)\left(2n-1+\sqrt{n(n-1)}\right)+\lambda\left(\sqrt n-\sqrt{n-1}\right)}\\ &=\dfrac{n+\sqrt{n(n-1)}}{2n+\sqrt{n(n-1)}-1+\lambda}\\ &=\dfrac 23+\dfrac{\sqrt{n(n-1)}-n+2(1-\lambda)}{6n+3\sqrt{n(n-1)}-3(1-\lambda)},\end{split}\]由于\[\dfrac 12<n-\sqrt{n(n-1)}\leqslant 1,\]于是分别取 $\lambda =\dfrac 12,\dfrac 34$，则有\[ \dfrac 23\left(n+\dfrac 12\right)\sqrt n\leqslant \sum_{k=1}^n\sqrt k <\dfrac 23\left(n+\dfrac 34\right)\sqrt n,\]因此原命题得证．