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题:计算 \(\displaystyle\lim_{n\to\infty}\sqrt{\small 4+\sqrt{4^2+\sqrt{\cdots+\sqrt{4^n}}}}\)Uu]O/ 解:令\(\;a_n=\sqrt{\small 4+\sqrt{4^2+\sqrt{\cdots+\sqrt{4^n}}}}.\) 易见MLL \(\qquad 3=\sqrt{4+\sqrt{4^2}+1}=\sqrt{\small 4+\sqrt{4^2+\sqrt{4^3}+1}}\)=Zi$XC \(\qquad\cdots=\sqrt{4+\sqrt{4^2+\sqrt{\cdots\sqrt{4^{n-1}+\sqrt{4^n}+1}}}}> a_n\)'RnE \(\quad\;\)令$\;\,b_{n,0} = \sqrt{4^n}+1,\;c_{n,0}=\sqrt{4^n}$`o7Vh $\qquad\,\;\; b_{n,k+1}=\sqrt{{\small 4^{n-k-1}+}b_{n,k}},\;c_{n,k+1}=\sqrt{{\small 4^{n-k-1}+}c_{n,k}}$ Z_^ 6h \(\because\;\;(b_{n,n-2},c_{n,n-2})=(3,a_n),\;b_{n,k+1}-c_{n,k+1}< \underset{\tiny\,}{\frac{\Large b_{n,k}-c_{n,k}}{\large 2^{n-k}}}\)7 \(\therefore\;\; 0< 3-a_n = b_{n,n-2}-c_{n,n-2}< 2^{-(2+3+\cdots+n)}\to 0.\quad\small\square\)p{-+
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