## Converging Root

#### 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.