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\[
n=\sqrt{n^2}
=\sqrt{1+(n-1)(n+1)}\\
n=a^{(n-1)}_{m}(m\in\mathbb{Z}_{>0}, n\in\mathbb{Z}_{>1})\\
b_{n-1}=\lim_{m\rightarrow\infty}a^{(n-1)}_{m}=\lim_{m\rightarrow\infty}n=n\\
\sqrt{ 1+2\sqrt{ 1+3\sqrt{ 1+\cdots } } }=b_2=3
\]

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