Question

Urgent va rog ex 19................

Answer 1

$\displaystyle Notez~t=\frac{1}{2}~si~voi~avea~a_n=(1+t) \left( 1+t^2 \right) \left( 1+t^{2^2} \right) ... \left(1+t^{2^n} \right). \\ \\ Bazat~pe~faptul~ca~ \left( 1+x^{2^n} \right) \left( 1-x^{2^n} \right )=1- \left(x^{2^n} \right)^2=1-x^{2 \cdot 2^n}= \\ \\ = 1-x^{2^{n+1}},~il~voi~inmulti~si~imparti~pe~a_n~cu~1-t. \\ \\ Rezulta:$

$\displaystyle a_n= \frac{(1-t)(1+t)(1+t^2) \left(1+t^{2^2} \right)... \left(1+t^{2^n} \right)}{1-t}= \\ \\ = \frac{(1-t^2)(1+t^2) \left(1+t^{2^2} \right)... \left(1+t^{2^n} \right)}{1-t}= \\ \\ =\frac{\left(1-t^{2^2} \right) \left( 1+t^{2^2}\right)...\left(1+t^{2^n} \right)}{1-t}= \\ \\= \frac{\left(1-t^{2^3} \right)... \left( 1+t^{2^n} \right)}{1-t}= \\ \\ =...= \\ \\ = \frac{\left(1-t^{2^n} \right) \left(1+t^{2^n} \right)}{1-t}= \\ \\ = \frac{1-t^{2^{n+1}}}{1-t}.$

$\displaystyle Cum~t \in (0,1)~\Rightarrow~t^{2^{n+1}} \to 0,~deci~a_n \to \frac{1}{1-t}. \\ \\ Deci~limita~este~\frac{1}{1-t}=2.$