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易知 $\sqrt{a}+\sqrt{b}>\sqrt {a+b}$ ($a,b>=1$) 则 $\sqrt{1+\sqrt{2+\sqrt{3...}}}<\sqrt{1}+\sqrt{\sqrt{2+\sqrt{3...}}}$ 同理 $\sqrt{1+\sqrt{2+\sqrt{3}}}<\sqrt{1}+\sqrt{\sqrt{2}}+\sqrt{\sqrt{\sqrt{3}}}+...=1^{\frac{1}{2}}+2^{\frac{1}{4}}+3^{\frac{1}{8}}+4^{\frac{1}{16}}...$ 因为 $1^{\frac{1}{2}}+2^{\frac{1}{4}}+3^{\frac{1}{8}}+4^{\frac{1}{16}}..\approx1++3^{\frac{1}{8}}+4^{\frac{1}{16}}..$