Let a epsilon 0 1 Prove that any sequence xn of real numbers

Let a epsilon (0, 1). Prove that any sequence (x_n) of real numbers satisfying the recurrence relation. X_n+1 = alpha x_n + (1 - alpha) x_n-1 has a limit and find an expression for the limit in terms of alpha, x_0 and x_1.

Solution

Given that n+1th term of the sequence depends on n-1th term and nth term by the following

recurrene relation

x_n+1=a x_n-1+(1-a) X_n

Since a belongs to (0,1) 1-a also must belong to (0,1)

So a x_n-1+(1-a) X_n< x_n-1+ X_n

As a result

x_n+1 < x_n-1+ X_n< x_n-3+ X_n-2+ x_n-2+ X_n-1

Thus the sequence is a decreasing sequence and since a and 1-a are less than 1,

xn converges to 0

a converges to 0.5