Let

Since a_1=3, solving the above equation gives

\alpha = 3+2\sqrt{2}

(cancel negative result since a_n>0 )

Substituting the value of prior terms in terms of \alpha the recursive formula gi

a_2=2a^2_1-1=2(\dfrac{1}{2}(\alpha +\dfrac{1}{\alpha } )^2-1

=\dfrac{1}{2}(\alpha^2 +\dfrac{1}{\alpha ^2 } )

a_3=\dfrac{1}{2}(\alpha ^4+\dfrac{1}{\alpha ^4 } )

\dots

a_n=\dfrac{1}{2}(\alpha ^{2^{n-1}}+\dfrac{1}{\alpha ^{2^{n-1}} } )

Then,

a_1\cdotp a_2\dots a_{n-1}

=\dfrac{1}{2^{n-1}}(\alpha +\dfrac{1}{\alpha })(\alpha^2 +\dfrac{1}{\alpha ^2 })\dots(\alpha ^{2^{n-2}}+\dfrac{1}{\alpha^{2^{n-2} } })

=\dfrac{1}{2^{n-1}\cdotp ( \alpha -\dfrac{1}{\alpha } )}(\alpha ^2-\dfrac{1}{\alpha ^2 }) (\alpha ^2+\dfrac{1}{\alpha^2 })\dots (\alpha ^{2^{n-2}}+\dfrac{1}{\alpha ^{2^{n-2}} } )

=\dfrac{\alpha ^{2^{n-1}}-\dfrac{1}{\alpha ^{2^{n-1}} } }{2^{n-1}\cdotp (\alpha -\dfrac{1}{\alpha } ) }

Therefore,

\lim\limits_{n\to \infty}\dfrac{a_n}{2^na_1a_2\dots a_{n-1}}

=\lim\limits_{n\to \infty}\dfrac{1}{4}\cdotp \dfrac{(\alpha -\dfrac{1}{\alpha } )a_n}{\alpha ^{2n-1}-\dfrac{1}{\alpha ^{2n-1} } }

=\lim\limits_{n\to \infty}\dfrac{1}{4}\cdotp \dfrac{(\alpha -\dfrac{1}{\alpha } )\dfrac{1}{2}(\alpha ^{2^{n-1}}+\dfrac{1}{\alpha ^{2^{n-1}} } )}{\alpha ^{2n-1}-\dfrac{1}{\alpha ^{2n-1} } }

=\dfrac{1}{4}\cdotp (\alpha -\dfrac{1}{\alpha })

=\sqrt{2}