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Problem

Consider the iterative formula, $u$_$n$+1 = $u$_$n$ + 2. By investigating the behaviour it should become clear that all positive starting values converge to the limit 4.

What form must the positive integer, $m$, take, for the iterative formula, $u$_$n$+1 = $u$_$n$ + $m$, to converge to an integral root?

Solution

We shall assume that the limit, $L$, exists, such that for sufficiently large values of $n$, $u$_$n$+1 $u$_$n$, and at limit, $u$_$n$+1 = $u$_$n$ = $L$.

Therefore, $L$ + $m$ = $L$, $L$ = $L$ $m$.

Squaring both sides, $L$ = $L$² 2$mL$ + $m$², and $L$² (2$m$+1)$L$ + $m$² = 0.

Using the quadratic formula, $L$ = (2$m$+1)(4$m$+1)/2.

As 2$m$+1 is odd and 4$m$+1 will be an odd square, their sum will be even, so it will always be divisible by 2. That is, the only condition is that 4$m$+1 is a perfect square.

Writing, 4$m$+1 = $k$², $m$ = ($k$² 1)/4, it is clear that $k$ must be odd,
let $k$ = 2$a$ 1.

$m$ = ((4$a$ 4$a$ + 1) 1)/4 = $a$($a$ 1).

Hence, if $m$ is of the form, $a$($a$ 1), the iterative form will converge to an integral root.

The solution, $L$ = (2$m$+1)(4$m$+1)/2, suggests that there are two roots of convergence. Prove that the iterative formula will always converge to the upper limit for all positive starting values.