Limit of a sequence of fractionsI have that: S n = b ⋅ S n...

aanpalendmw

aanpalendmw

Answered

2022-07-17

Limit of a sequence of fractions
I have that:
S n = b S n 1 b + S n 1 + a
where n N + and S 1 = a + b
So, for S 2 we will get:
S 2 = b S 2 1 b + S 2 1 + a = b S 1 b + S 1 + a = b ( a + b ) b + ( a + b ) + a
And for S 3 :
S 3 = b S 3 1 b + S 3 1 + a = b S 2 b + S 2 + a = b ( b ( a + b ) b + ( a + b ) + a ) b + ( b ( a + b ) b + ( a + b ) + a ) + a
And so on. Now my question is: what happens when n ?

Answer & Explanation

Mira Spears

Mira Spears

Expert

2022-07-18Added 14 answers

Assume that S n = p n q n is associated with ( p n , q n ) R 2 . The recurrence
(1) S n = ( a + b ) S n 1 + a b S n 1 + b
can be written in the following form
(2) ( p n q n ) = ( a + b a b 1 b ) ( p n 1 q n 1 )
hence the closed form of both the { p n } n 0 and the { q n } n 0 sequences just depends on the diagonalization of M = ( a + b a b 1 b ) , whose eigenvalues are given by ζ ± = ( a + 2 b ) ± a 2 + 4 a b 2 .
In particular
(3) S n = A ζ + n + B ζ n C ζ + n + D ζ n
for some constants ( A , B , C , D ) depending on the initial values. By taking the limit as n + , we simply get A C . On the other hand, by assuming S n L as n , such limit has to fulfill
(4) L = a + b L b + L
hence the only chances are L = 1 2 ( a ± a 2 + 4 a b )

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