Đặt [tex]\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}=a(a>0)\Rightarrow 2+\sqrt{2+\sqrt{2+\sqrt{2}}}=a^{2}\Leftrightarrow \sqrt{2+\sqrt{2+\sqrt{2}}}=a^{2}-2[/tex]
Có [tex]\sqrt{2}<2\Rightarrow \sqrt{2+\sqrt{2}}<\sqrt{2+2}=2\Rightarrow \sqrt{2+\sqrt{2+\sqrt{2}}}< \sqrt{2+2}=2\Rightarrow \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}< \sqrt{2+2}=2\Leftrightarrow a<2[/tex]
Lại có: $2<2+\sqrt{2+\sqrt{2+\sqrt{2}}}\Rightarrow\sqrt{2}<a\Rightarrow1<a$
Suy ra [TEX]1<a<2[/TEX]
Khi đó, [tex]H=\frac{2-a}{2-(a^{2}-2)}=\frac{2-a}{4-a^{2}}=\frac{1}{2+a}[/tex]
Có [TEX]1<a<2\Rightarrow 3<a+2<4\Rightarrow \frac{1}{4}<\frac{1}{a+2}<\frac{1}{3}\Leftrightarrow \frac{1}{4}<H<\frac{1}{3}(dpcm)[/TEX]