Woah. It's not everyday I can bring out the Banach fixed point theorem. The theorem essentially will give sufficient conditions for when OP’s function will converge and perhaps explain why his function fails to converge for values roughly greater than 15.

Setting Up Notation First going to define the sequence [;\phi{t} = \lambda^{-\phi{t-1}};] for [;t \geq 1;] and [; \phi_{0}=k;] with the constraints that [; \lambda \geq 1;]. In OP's post: [;k= \frac{1}{\lambda};] [;\phi(\lambda)= \lim{t\to\infty} \phi{t};] when such limit exists. *Note the LHS is OP's notations and the RHS is my notation.

Now we can define [;f(x) = \lambda^{-x};] such that [;\phi{t} = f(\phi{t-1});]

Proving Conditions Necessary to Apply Theorem Now observe that [;f(x);] is a monotonically decreasing and continuously differentiable over [;\mathbb{R};] with

[;f^{\prime} (x)= -\lambda^{{-x}ln(\lambda);]} and [;f^{\prime}(x) \leq 0, \forall x\in \mathbb{R};] since [;ln(\lambda)\geq 0, \forall \lambda \geq 1;]

To apply the Banach Fixed Point theorem we need to show that [;f(x);] is a contractive mapping on some bounded interval. On such interval we will require that [;|f^{\prime}(x)| < 1;] which will provide the necessary conditions to apply the theorem.

[;|f^{\prime}(x)| <1 \implies \lambda^{{-x}ln(\lambda)} <1;]

[; \imples ln(ln(\lambda)) < x ln(\lambda);]

Thus we require [;x> \frac{ln(ln(\lambda))}{ln(\lambda)};] and if we set our other endpoint to be [;f(\frac{ln(ln(\lambda))}{ln(\lambda)});]

Let [;a = \frac{ln(ln(\lambda))}{ln(\lambda)};] and let [;b =f(\frac{ln(ln(\lambda))}{ln(\lambda)});] It can be easily verified that [;f: [a,b] \mapsto [a,b];] and [; \forall x,y \in [a,b], |f(a) - f(b)| < |a-b|;] Thus we can now apply the Banach fixed point theorem to state that [; lim{t \to \infty} \phi{t};] exists and is equal to the unique value [;x^{\star};] such that [;f(x^{{\star})=x{\star};]} Thus predicting that OP’s [; \phi( \lambda) = k^{x{\star}};] This of course assumes that [;a<b;] which is not true in general. This is only true for [; \lambda \in [1, e^{e}];] [;e^{e} = 15.15 … ;] And thus might explain why OP’s method failed to converge for values such as 16.