设级数
$$\sum_{n=1}^{\infty}2^{-n}a_{n+1}(a_{1}a_{2}\cdots a_{n}a_{n+1})^{-\frac{1}{2}}$$
收敛，证明数列
$$x_{n}=\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}$$
也收敛，这里$a_{1},a_{2},\cdots,a_{n},\cdots$为正实数。

证明：注意到$\{x_{n}\}$是单调递增的。而又有
$$\begin{align*} &x_{n+1}-x_{n}\\ &=\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}-\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}\\ &=\frac{\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}-\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}+\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)}{\left(\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}}+\sqrt{a_{1}+\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}} \right)}\\ &<\frac{\sqrt{a_{2}+\cdots+\sqrt{a_{n+1}}}-\sqrt{a_{2}+\cdots+\sqrt{a_{n}}}}{2\sqrt{a_{1}}}\\ &<\frac{\sqrt{a_{3}+\cdots+\sqrt{a_{n+1}}}-\sqrt{a_{3}+\cdots+\sqrt{a_{n}}}}{2\sqrt{a_{1}}\cdot 2\sqrt{a_{2}}}\\ &<\cdots\\ &<\frac{\sqrt{a_{n+1}}}{2^{n}\sqrt{a_{1}a_{2}\cdots a_{n}}}\\ &=\frac{a_{n+1}}{2^{n}\sqrt{a_{1}a_{2}\cdots a_{n+1}}} \end{align*}$$
而由于级数收敛，所以数列$\{x_{n}\}$收敛。