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Problem

A second order recurrence relation is defined by $u$_n+1 = ($u$_n+$u$_n1)/2; that is, each new term is the mean of the previous two terms.

For example, when $u$₁=2 and $u$₂=5, we generate the following sequence:

2, 5, 3.5, 4.25, 3.875, ... .

Find the limit of the sequence for different starting values.

Solution

Consider the following diagram.

By the definition of the sequence, the next term, $u$_k+1 will lie between the two previous terms, $u$_k and $u$_k1. Thus, as this process continues, each iteration will reduce the upper and lower bounds, demonstrating that the sequence converges towards a value.

Let the distance between the first and second term be defined as one unit, so that, from the first term $u$₁, the terms follow the following series:

S = 1 1/2 + 1/4 1/8 + ...
2S = 2 1 + 1/2 1/4 + 1/8 ...
3S = 2 S = 2/3

That is, the sequence converges towards a value, L, which is 2/3 above the first term: L = $u$₁+2($u$₂$u$₁)/3 = ($u$₁+2$u$₂)/3.

If $u$₁ $u$₂ then the sequence follows the series:

S = -1 + 1/2 1/4 + ...
2S = -2 + 1 1/2 + 1/4 ...
3S = -2 S = -2/3

And we get the same limit, L = $u$₁2($u$₁$u$₂)/3 = ($u$₁+$u$₂)/3.

What about the third order recurrence relation: $u$_n+1 = ($u$_n+$u$_n1+$u$_n2)/3 ?
Investigate the $k$ th order recurrence relation.