# Convergence of a Non-linear Recursive Sequence with Fractional Exponent

I am exploring a recursively defined sequence involving non-integer exponents, and I aim to find or confirm the limit of $$\left\{b_n\right\} \text { as } n \rightarrow \infty \text {. }$$ The sequence is defined as follows:

Let $$x_1, x_2, \ldots, x_t$$ be initial fixed positive real numbers, and define the sequence $$\{b_n\}$$ by:

• For $$n \leq t$$, $$b_n = x_n$$.
• For $$n > t$$, $$b_n = b_{n-1}^p + b_{n-2}^p + \cdots + b_{n-k}^p,$$ where $$p$$ is a real number with $$0 < p < 1$$.

### Progress and Approach:

Given the concavity of the function $$f(x) = x^p$$ for $$x > 0$$ and $$0 < p < 1$$, we observe that $$f(x) \leq x$$ for $$x \geq 1$$. This suggests that each term in the sequence contributes progressively less as it grows larger, potentially indicating boundedness or stabilization of $$\{b_n\}$$.

Assuming the sequence might converge to a limit $$L$$, we can expect $$L$$ to satisfy the steady-state condition: $$L = k L^p.$$ Solving this leads to: $$L = k^{1/(1-p)}.$$ This suggests a fixed point, which we hypothesize might act as an attractor for the sequence.

To further validate this, I considered the function $$f : [0, \infty)^k \to [0, \infty)$$ defined by: $$f(x_1, \ldots, x_k) = x_1^p + \ldots + x_k^p.$$ Given that the derivative $$\frac{d}{dx}(x^p) = p x^{p-1}$$ is less than 1 for all $$x > 1$$ and $$p < 1$$, I aimed to examine if $$f$$ acts as a contraction mapping under a suitable metric. Defining the metric as: $$d((x_1, \ldots, x_j), (y_1, \ldots, y_j)) = \max |x_i - y_i|,$$ one needs to show that: $$|f(x_1, \ldots, x_j) - f(y_1, \ldots, y_j)| \leq C \max |x_i - y_i|,$$ where $$C < 1$$. This would imply that $$f$$ gradually brings terms closer together, suggesting convergence towards the fixed point $$L$$.

Initially, setting $$M = \max(x_1, \ldots, x_t)$$, it leads to: $$b_{t+1} \leq k M^p$$ and following the recurrence: $$b_{t+2} \leq k \cdot (k M^p)^p = k^{1+p} M^{p^2}$$ indicating each term might be bounded by a decreasing sequence as $$M^{p^n}$$ approaches zero for large $$n$$, given $$p^n$$’s exponential decay.

Considering these, how can I rigorously prove the hypothesized convergence towards $$L = k^{1/(1-p)}$$, or assess how the sequence will converge/bound instead? I would greatly appreciate any insights, corrections, or suggestions on methods to rigorously confirm whether $$\{b_n\}$$ converges and to what limit.

• A small note: you distinguish $t$ from $k$, but that doesn't really increase generality, since we must have $t \ge k$ for the sequence to be well-defined, and if $t > k$ then the first $t - k$ terms of the sequence have no bearing on the rest. May 13 at 16:49
• @crb233 That's correct, but I am afraid I don't understand what conclusions you end up with this. Does it mean $t$ is dependent on $k$ or what exactly? May 13 at 17:11
• We can simplify the recurrent formula as follows. We assume that $t\ge k$ to define the sequence well, as was noted by crb233. Let $n>t$. Then $$b_n = b_{n-1}^p + b_{n-2}^p + \cdots + b_{n-k}^p,$$ $$b_{n+1} = b_{n}^p + b_{n-1}^p + \cdots + b_{n-k+1}^p=b_{n}^p+b_n-b_{n-k}^p.$$ May 13 at 19:55
• @AlexRavsky Yes, that's true. But I am afraid I do not know to understand how to find a limit for this recurrence after having that. May 14 at 1:51
• @crb233 If there is any progress you think deserves attention, I would be very glad to see them. May 14 at 12:52

Following crb233's comment we assume that $$t\ge k$$. Put $$K=k^{1/(1-p)}$$, $$L_{-}=\liminf_{n\to\infty} b_n:=\lim_{n\to\infty}\inf_{k\ge n} b_k\mbox{ and } L_+=\limsup_{n\to\infty} b_n:=\lim_{n\to\infty}\sup_{k\ge n} b_k$$
Let $$c_1>0$$ be any number and $$(c_n)_{n\in\mathbb N}$$ be the sequence of real numbers such that $$c_{n+1}=c_n^p$$ for each natural $$n\ge 2$$. It is easy to see that $$\lim_{n\to\infty} c_n=1$$. Since the numbers $$b_1,b_2,\dots,b_t$$ are positive and for each natural $$n>t$$ we have $$b_n\ge b_{n-1}^p$$, we obtain that $$L_-\ge 1$$. Moreover, the recurrence for the sequence $$(b_n)_{n\in\mathbb N}$$ implies that $$L_-\ge k(L_-)^p$$, so $$L_-\ge K$$.
Now let $$M=\max\{b_1,\dots,b_k,K\}$$. Then $$kM^p\le M$$, so $$b_n\le M$$ for each natural $$n$$. Moreover, the recurrence for the sequence $$(b_n)_{n\in\mathbb N}$$ implies that $$L_+\le k(L_+)^p$$, so $$L_+\le K$$.
Thus $$K=L_-=L_+=L=\lim_{n\to\infty} b_n$$ which proves the desired answer.