\[a_{n+1} = a_n + a_{n-1}\]
with \[a_0 =a_1 =1\]
. In English we add two terms to get the next term.The ratio of successive Fibonacci numbers tends to a limit
\[\frac{a_{n+1}}{a+n} = \frac{ \sqrt{5} -1}{2}\]
as \[n \rightarrow \infty\]
Consider the 'generalized Fibonacci sequence' defined by the rule \[a_{n+1} =A a_n +B a_{n-1} , \: A, B > 0\]
Does the ration of successive terms to to a limit for this sequence? Yes it does. Divide the rule that generates this sequence by
\[a_n\]
to give\[\frac{a_{n+1}}{a_n} =A +B \frac{a_{n-1}}{a_n}\]
As
\[n \rightarrow \infty\]
, if \[\frac{a_{n+1}}{a_n} \rightarrow l\]
then \[\frac{a_n}{a_{n-1}} \rightarrow \frac{1}{l}\]
Then
\[\frac{a_{n+1}}{a_n} =A +B \frac{a_{n-1}}{a_n}\]
becomes \[l =A +\frac{B}{l}\]
Multiply by
\[l\]
and rearrange to obtain \[l^2 - Al -B=0\]
The solutions are
\[l= \frac{A \pm \sqrt{A^2 +4B}}{2}\]
The positive option gives the correct value of
\[l\]
. The negative option gives a negative vale. This is wrong since \[A,B >0\]
so \[A < \sqrt{A^2 +4B}\]
.