Recurring Styles #12 - Limit of a Recurrence!

Calculus Level 5

$\large{ S = \sum_{n=1}^\infty \dfrac{a_{2n+2}}{a^2_{n-1} a^2_{n+1}} }$

Let $(a_n)$ be a sequence defined by $a_0 = 1, a_1 = 2$ , and for $n \geq 2$ , $a_n = a_{n-1} + a_{n-2}$ . If $S$ can be expressed as $\dfrac{A}{B}$ where $A,B \in \mathbb Z^+, \ \gcd(A,B)=1$ . Submit the value of $A+B$ as your answer.

The answer is 9.

1 solution

Brian Charlesworth
Aug 31, 2015

We can first observe that the sequence $\{a_{n}\}$ is just the Fibonacci sequence with an index displacement of $2,$ (i.e., $a_{n} = F_{n+2}$ ), in which case we see that

$S = \displaystyle\sum_{n=3}^{\infty} \dfrac{F_{2n}}{F_{n-1}^{2}F_{n+1}^{2}}.$

Now $F_{2n} = F_{n+1}^{2} - F_{n-1}^{2},$ , (*), (proof below), so

$S = \displaystyle\sum_{n=3}^{\infty} \left(\dfrac{1}{F_{n-1}^{2}} - \dfrac{1}{F_{n+1}^{2}} \right),$

which is a telescoping series with terms canceling pairwise except for the first components of the first two bracketed terms. Thus

$S = \dfrac{1}{F_{2}^{2}} + \dfrac{1}{F_{3}^{2}} = \dfrac{1}{1^{2}} + \dfrac{1}{2^{2}} = \dfrac{5}{4}.$

We then see that $A + B = 5 + 4 = \boxed{9}.$

Proof of (*): Note first that

$F_{m+n} = F_{m+1}F_{n} + F_{m}F_{n-1},$

which can be proved by induction on $n.$ With $m = n$ we then have that

$F_{2n} = F_{n+1}F_{n} + F_{n}F_{n-1} = F_{n}(F_{n+1} + F_{n-1}) =$

$(F_{n+1} - F_{n-1})(F_{n+1} + F_{n-1}) = F_{n+1}^{2} - F_{n-1}^{2}.$

Recurring Styles #12 - Limit of a Recurrence!

The answer is 9.

1 solution

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