College Math

Algebra Level 5

$\large \prod_{k=0}^{\infty}\left(1-\frac1{a_k}\right)$

Consider the recurrence relation $a_k = a_{k-1} ^2 - 2$ with $a_0 = 5/2$ . If the product above can be stated in terms of $\dfrac AB$ , where $A$ and $B$ are coprime positive integers, find the value of $A+B$ .

The answer is 10.

1 solution

Kushal Dey
Dec 2, 2020

We need to observe 2 things 1. a(k)=a(k-1)²-2 => a(k)+1=(a(k-1)+1)(a(k-1)-1) => a(k-1)-1=(a(k)+1)/(a(k-1)+1), this gives us a telescopic product 2. a(0)=5/2=2+1/2, let c=2, this a(0)=c+1/c. a(1)=(c+1/c)²-2=c²+1/c², it can be easily shown that a(n)=c^(2^n)+1/c^(2^n). We can write our infinite product till a finite number n and then take limit n tending infinity. Thus we have Lt n-> inf ((a(n+1)+1)/(a(n)+1))×((a(n)+1)/a(n-1)+1)×...((a(1)+1)/(a(0)+1))/(a(0)a(1).....a(n)). Numerator simplifies to (a(n+1)+1)/(a(0)+1) telescopically(using eqn 1) For denominator we can use eqn 2 to write a(n)= (c^(2^(n+1))-1/c^(2^(n+1)))/(c^(2^n)-1/c^(2^n)) and it also telescopically simplifies to (c^(2^(n+1))-1/c^(2^(n+1)))/(c-1/c). So our product is Lt n-> inf ( (c-1/c)(c^(2^(n+1))+1+1/c^(2^(n+1)))/(c+1+1/c)(c^(2^(n+1))-1/c^(2^(n+1))) As n approaches infinity, 1 and 1/c^(2^(n+1)) are too small as compared to c^(2^(n+1)), so our product finally becomes (c-1/c)/(c+1+1/c). Put c=2, and the answer is 3/7.

College Math

The answer is 10.

1 solution

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