Tolaso J Kos wrote:Let \(a_k \) be a recursive sequence defined as \( \displaystyle a_0=\frac{5}{2} , \; a_k=a_{k-1}^2-2 ,\;\; k \in \mathbb{N} \).
Evaluate the product: $$\prod_{k=0}^{\infty} \left( 1-\frac{1}{a_k} \right) $$
My solution
Note that
$a_{0}=\frac{5}{2}=\frac{2^{2}+1}{2^{1}}=\frac{2^{2^{1}}+1}{2^{2^{0}}}$
$a_{1}=\left( \frac{5}{2} \right)^{2}-2=\frac{17}{4}=\frac{2^{2^{2}}+1}{2^{2^{1}}}$
$a_{2}=\left( \frac{17}{4} \right)^{2}-2=\frac{257}{16}=\frac{2^{2^{3}}+1}{2^{2^{2}}}$
$a_{3}=\left( \frac{257}{16} \right)^{2}-2=\frac{65537}{256}=\frac{2^{2^{4}}+1}{2^{2^{3}}}$
$a_{4}=\left( \frac{65537}{256} \right)^{2}-2=\frac{4294967297}{65536}=\frac{2^{2^{5}}+1}{2^{2^{4}}}$
$...$
$a_{N}=\frac{2^{2^{N+1}}+1}{2^{2^{N}}}$
$P=\prod\limits_{n=0}^{+\infty }{\left( 1-\frac{1}{a_{n}} \right)}=\left( 1-\frac{1}{a_{0}} \right)\left( 1-\frac{1}{a_{1}} \right)\left( 1-\frac{1}{a_{2}} \right)\cdot ...=\underset{N\to +\infty }{\mathop{\lim }}\,\left( 1-\frac{1}{a_{0}} \right)\left( 1-\frac{1}{a_{1}} \right)\left( 1-\frac{1}{a_{2}} \right)\cdot ...\cdot \left( 1-\frac{1}{a_{N}} \right)$
$=\underset{N\to +\infty }{\mathop{\lim }}\,\left( 1-\frac{2^{2^{0}}}{2^{2^{1}}+1} \right)\left( 1-\frac{2^{2^{1}}}{2^{2^{2}}+1} \right)\left( 1-\frac{2^{2^{2}}}{2^{2^{3}}+1} \right)\cdot ...\cdot \left( 1-\frac{2^{2^{N}}}{2^{2^{N+1}}+1} \right)$
$=\underset{N\to +\infty }{\mathop{\lim }}\,\left( \frac{2^{2^{1}}-2^{2^{0}}+1}{2^{2^{1}}+1} \right)\left( \frac{2^{2^{2}}-2^{2^{1}}+1}{2^{2^{2}}+1} \right)\left( \frac{2^{2^{3}}-2^{2^{2}}+1}{2^{2^{3}}+1} \right)\cdot ...\cdot \left( \frac{2^{2^{N+1}}-2^{2^{N}}+1}{2^{2^{N+1}}+1} \right)$
$=\underset{N\to +\infty }{\mathop{\lim }}\,\frac{\left( \left( 2^{2^{0}}+1 \right)\left( 2^{2^{1}}-2^{2^{0}}+1 \right) \right)\left( \left( 2^{2^{1}}+1 \right)\left( 2^{2^{2}}-2^{2^{1}}+1 \right) \right)\cdot ...\cdot \left( \left( 2^{2^{N}}+1 \right)\left( 2^{2^{N+1}}-2^{2^{N}}+1 \right) \right)}{\left( 2^{2^{0}}+1 \right)\left( 2^{2^{N+1}}+1 \right)\left( \left( 2^{2^{1}}+1 \right)\left( 2^{2^{2}}+1 \right)\cdot ...\cdot \left( 2^{2^{N}}+1 \right) \right)^{2}}$
$=\frac{1}{3}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{\left( 2^{3}+1 \right)\left( 2^{6}+1 \right)\left( 2^{9}+1 \right)\cdot ...\cdot \left( 2^{3\cdot 2^{N}}+1 \right)}{\left( 2^{2^{N+1}}+1 \right)\left( \left( 2^{2^{1}}+1 \right)\left( 2^{2^{2}}+1 \right)\cdot ...\cdot \left( 2^{2^{N}}+1 \right) \right)^{2}}$
$=\frac{1}{3}\cdot \frac{\left( 2^{2^{1}}-1 \right)^{2}}{2^{3}-1}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{2^{3\cdot 2^{N+1}}-1}{\left( 2^{2^{N+1}}+1 \right)\left( \left( 2^{2^{1}}-1 \right)\left( 2^{2^{1}}+1 \right)\left( 2^{2^{2}}+1 \right)\cdot ...\cdot \left( 2^{2^{N}}+1 \right) \right)^{2}}$
$=\frac{3}{7}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{2^{3\cdot 2^{N+1}}-1}{\left( 2^{2^{N+1}}+1 \right)\left( 2^{2^{N+1}}-1 \right)^{2}}=\frac{3}{7}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{\left( 2^{2^{N+1}} \right)^{3}-1}{\left( 2^{2^{N+1}}+1 \right)\left( 2^{2^{N+1}}-1 \right)^{2}}$
$=\frac{3}{7}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{\left( 2^{2^{N+1}}-1 \right)\left( 2^{2^{N+1}\cdot 2}+2^{2^{N+1}}+1 \right)}{\left( 2^{2^{N+1}}+1 \right)\left( 2^{2^{N+1}}-1 \right)^{2}}=\frac{3}{7}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{2^{2^{N+2}}+2^{2^{N+1}}+1}{2^{2^{N+2}}-1}=\frac{3}{7}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{1+\frac{1}{2^{2^{N+2}-2^{N+1}}}+\frac{1}{2^{2^{N+2}}}}{1-\frac{1}{2^{2^{N+2}}}}$
$=\frac{3}{7}\underset{N\to +\infty }{\mathop{\lim }}\,\frac{1+\frac{1}{2^{2^{N+1}}}+\frac{1}{2^{2^{N+2}}}}{1-\frac{1}{2^{2^{N+2}}}}=\frac{3}{7}\cdot \frac{1+0+0}{1-0}=\frac{3}{7}$
Done, podrias proponer tu solucion ?
Saludos cordiales ^^